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EUCLID'S ELEMENTS OF GEOMETRY 



The Greek text of J.L. Heiberg (1883-1885) 

from Euclidis Elementa, edidit et Latine interpretatus est I.L. Heiberg, in aedibus 

B.G. Teubneri, 1883-1885 



edited, and provided with a modern English translation, by 

Richard Fitz pat rick 



First edition - 2007 

Revised and corrected - 2008 

ISBN 978-0-6151-7984-1 



Contents 



Introduction 4 

Book 1 5 

Book 2 49 

Book 3 69 

Book 4 109 

Book 5 129 

Book 6 155 

Book 7 193 

Book 8 227 

Book 9 253 

Book 10 281 

Book 11 423 

Book 12 471 

Book 13 505 

Greek-English Lexicon 539 



Introduction 



Euclid's Elements is by far the most famous mathematical work of classical antiquity, and also has the distinction 
of being the world's oldest continuously used mathematical textbook. Little is known about the author, beyond 
the fact that he lived in Alexandria around 300 BCE. The main subjects of the work are geometry, proportion, and 
number theory. 

Most of the theorems appearing in the Elements were not discovered by Euclid himself, but were the work of 
earlier Greek mathematicians such as Pythagoras (and his school), Hippocrates of Chios, Theaetetus of Athens, and 
Eudoxus of Cnidos. However, Euclid is generally credited with arranging these theorems in a logical manner, so as to 
demonstrate (admittedly, not always with the rigour demanded by modern mathematics) that they necessarily follow 
from five simple axioms. Euclid is also credited with devising a number of particularly ingenious proofs of previously 
discovered theorems: e.g., Theorem 48 in Book 1. 

The geometrical constructions employed in the Elements are restricted to those which can be achieved using a 
straight-rule and a compass. Furthermore, empirical proofs by means of measurement are strictly forbidden: i.e., 
any comparison of two magnitudes is restricted to saying that the magnitudes are either equal, or that one is greater 
than the other. 

The Elements consists of thirteen books. Book 1 outlines the fundamental propositions of plane geometry, includ- 
ing the three cases in which triangles are congruent, various theorems involving parallel lines, the theorem regarding 
the sum of the angles in a triangle, and the Pythagorean theorem. Book 2 is commonly said to deal with "geometric 
algebra", since most of the theorems contained within it have simple algebraic interpretations. Book 3 investigates 
circles and their properties, and includes theorems on tangents and inscribed angles. Book 4 is concerned with reg- 
ular polygons inscribed in, and circumscribed around, circles. Book 5 develops the arithmetic theory of proportion. 
Book 6 applies the theory of proportion to plane geometry, and contains theorems on similar figures. Book 7 deals 
with elementary number theory: e.g., prime numbers, greatest common denominators, etc. Book 8 is concerned with 
geometric series. Book 9 contains various applications of results in the previous two books, and includes theorems 
on the infinitude of prime numbers, as well as the sum of a geometric series. Book 10 attempts to classify incommen- 
surable {i.e., irrational) magnitudes using the so-called "method of exhaustion", an ancient precursor to integration. 
Book 11 deals with the fundamental propositions of three-dimensional geometry. Book 12 calculates the relative 
volumes of cones, pyramids, cylinders, and spheres using the method of exhaustion. Finally, Book 13 investigates the 
five so-called Platonic solids. 

This edition of Euclid's Elements presents the definitive Greek text — i.e., that edited by J.L. Heiberg (1883- 
1885) — accompanied by a modern English translation, as well as a Greek-English lexicon. Neither the spurious 
books 14 and 15, nor the extensive scholia which have been added to the Elements over the centuries, are included. 
The aim of the translation is to make the mathematical argument as clear and unambiguous as possible, whilst still 
adhering closely to the meaning of the original Greek. Text within square parenthesis (in both Greek and English) 
indicates material identified by Heiberg as being later interpolations to the original text (some particularly obvious or 
unhelpful interpolations have been omitted altogether) . Text within round parenthesis (in English) indicates material 
which is implied, but not actually present, in the Greek text. 

My thanks to Mariusz Wodzicki (Berkeley) for typesetting advice, and to Sam Watson & Jonathan Fenno (U. 
Mississippi), and Gregory Wong (UCSD) for pointing out a number of errors in Book 1. 



4 



ELEMENTS BOOK 1 

Fundamentals of Plane Geometry Involving 

Straight-Lines 



5 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



"Opoi. 

a'. Er^elov eoxiv, ou ^epoc; ou'dev. 
P'. rpa^rj 8s nfjxoc; a7tXocxe<;. 
y'. rpa^urj<; Be Ttepaxoc or]^eTa. 

8'. EO'deToc ypo^iuf| eoxiv, ffiic, 15 taou tou; ecp' eauxrji; 
ar)fie[oi<; xelxai. 

e'. 'Eracpdveioc Be eoxiv, o urjxoc; xdi rcXdxoc; [iovov e/ei. 

<r'. Ti/iucpavdac; 8s Ttepaxa ypajjijiai. 

C- 'EtutteSoc; emcpdveid eoxiv, fjxic; ioou xaig ecp' 
eauxrjc; euT&eioac; xelxai. 

T)'. 'EtutccBoc; Be ytovia eaxiv f] ev sniTieScp Buo ypaji^Gv 
dnxo^ieviov dXXrjXwv xal u/] en' eui5eiac; xeijievtov npoc; 
dXXrjXac; xfiv ypa^cov x^ou;. 

"Oxav Be ai nspis/ouoai xr]v ywviav ypajjijidi eu-delai 
Goiv, eMuypaji^ioc; xaXeTxai f] ywvia. 

i'. "Oxav Be euiikTa in' eui&eTav oxa^eloa xdg ecpe^rjc; 
ywviac; Toac; dXXfjXaic; Ttoifj, 6p$r) exaxepa xfiv Tacov yioviCSv 
eoxi, xdi f) ecpeaxr)xuia eui9eTa xdi^exoc; xaXeTxai, ecp' fjv 

£(p£aXT)X£V. 

ia'. AjipXeTa ycovia eoxiv f] jiei^cjv op-drjc;. 

iP'. 'Ocelot 8s: f] eXdaocov op^rjc;. 

iy'. "Opoc; eoxiv, o xivoc; eoxi Ttepotc;. 

18'. Sx^K a ^ OTl TO UIto Tlvo ? ^ xivwv opwv Ttepiexo^ievov. 

is'. KuxXoc; eoxi ox/jjia emiteBov Otto jiidc; ypajiurjc; 
TiEpiExojisvov [f] xaXsTxai nepicpepeia], npbc, fjv dcp' evoc; 
or^eiou xfiv evxoc; xou ax^axog xei^ieviov Ttdoai di 
npooTUTixouoai eu-delai [npog xrjv xoO xuxXou rcepicpepeiav] 
lacci dXXrjXaic; eiaiv. 

i=r'. Kevxpov 8s xou xuxXou xo or^eTov xaXeixai. 

Aidjiexpog 8s xou xuxXou eaxiv euaSe'id xic; Std xou 
xevxpou yjy^evr] xdi Tiepaxoujjievr) ecp' exdxepa xd jiepr) 
utto xrjc; xou xuxXou nepicpepeiac;, fjxic; xdi Si/a xeuvei xov 
xuxXov. 

it]'. 'HjiixuxXiov Be eoxi xo Ttepiex6|ievov ox^a Otto xe 
xrjc Biajjixpou xdi xrjg duoXa(jipavo|jievr]^ On' auxfjc; nepi- 
9epeiag. xevxpov 8e xou f)^ixuxX[ou xo auxo, o xdi xou 
xuxXou eaxiv. 

it}'. S)(r]jiaxa eui9uypaji|jid eoxi xd utto eu-deifiv rce- 
pie/o^ieva, xpiicXeupa ^iev xd uno xpiwv, xexpdnXeupa Se xd 
utio xeoodpwv, noXunXeupa Se xd uno tiXciovcov f\ xeaadpwv 
eu-deifiv nepiexo^ieva. 

x'. TGv Be xpiTiXeupwv o/iqtidxwv iaoTiXeupov [ie\ 
xpiycovov eoxi xo xdg xpelg Taac; exov TiXeupdc;, ioooxeXec; 
8e xo xdg 8uo [lovolc, Icolc, exov TtXeupdc;, oxaXrjvov 8e xo 
xdc; xpelg dviooug exov nXeupdc;. 

xa' "Exi 8e x«v xpiTiXeupwv oxr)|jidxwv 6pi9oycoviov [iev 
xpiycovov eoxi xo exov op'driv ycoviav, djipXuycjviov 8e xo 
exov djipXelav ywviav, o^uywviov 8e xo xdg xpelc; o^eiac; 
exov ywvia^. 



Definitions 

1. A point is that of which there is no part. 

2. And a line is a length without breadth. 

3. And the extremities of a line are points. 

4. A straight-line is (any) one which lies evenly with 
points on itself. 

5. And a surface is that which has length and breadth 
only. 

6. And the extremities of a surface are lines. 

7. A plane surface is (any) one which lies evenly with 
the straight-lines on itself. 

8. And a plane angle is the inclination of the lines to 
one another, when two lines in a plane meet one another, 
and are not lying in a straight-line. 

9. And when the lines containing the angle are 
straight then the angle is called rectilinear. 

10. And when a straight-line stood upon (another) 
straight-line makes adjacent angles (which are) equal to 
one another, each of the equal angles is a right-angle, and 
the former straight-line is called a perpendicular to that 
upon which it stands. 

11. An obtuse angle is one greater than a right-angle. 

12. And an acute angle (is) one less than a right-angle. 

13. A boundary is that which is the extremity of some- 
thing. 

14. A figure is that which is contained by some bound- 
ary or boundaries. 

15. A circle is a plane figure contained by a single line 
[which is called a circumference], (such that) all of the 
straight-lines radiating towards [the circumference] from 
one point amongst those lying inside the figure are equal 
to one another. 

16. And the point is called the center of the circle. 

17. And a diameter of the circle is any straight-line, 
being drawn through the center, and terminated in each 
direction by the circumference of the circle. (And) any 
such (straight-line) also cuts the circle in half.^ 

18. And a semi-circle is the figure contained by the 
diameter and the circumference cuts off by it. And the 
center of the semi-circle is the same (point) as (the center 
of) the circle. 

19. Rectilinear figures are those (figures) contained 
by straight-lines: trilateral figures being those contained 
by three straight-lines, quadrilateral by four, and multi- 
lateral by more than four. 

20. And of the trilateral figures: an equilateral trian- 
gle is that having three equal sides, an isosceles (triangle) 
that having only two equal sides, and a scalene (triangle) 
that having three unequal sides. 



6 



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ELEMENTS BOOK 1 



xp'. Twv Se xexpaTiXeupwv axr^dxtov xexpdyovov (iev 
eoxiv, 8 EaoTtXeupov xe eaxi xal op-doywviov, exepo^xet; 
8e, 8 op-doytoviov ^lev, oux looTtXeupov Se, po^poc; 8e, 8 
iooTtXeupov jiev, oux opiDoycoviov 86, po^poeiBei; 8s xo xdc; 
dmevavxiov TiXeup&c xe xal ywvtac laac, dXXr|Xoa<; £X ov j ° 
ouxe taoTiXeupov eaxiv oOxs op'Ooycoviov xd 8e itapd xauxa 
xexpaitXeupa xpane^ia xaXeurdco. 

xy'. IIapdXXr)Xo[ rioiv eii/dsToci, atxivec; ev tw auxw 
eraTieBo ouoai xal expaXXo^tevai sic; aTteipov ecp' exdxepa 
xd [lepr] era ^tr]8exepa au^mrnxouoiv dXXr]Xai<;. 



21. And further of the trilateral figures: a right-angled 
triangle is that having a right-angle, an obtuse-angled 
(triangle) that having an obtuse angle, and an acute- 
angled (triangle) that having three acute angles. 

22. And of the quadrilateral figures: a square is that 
which is right-angled and equilateral, a rectangle that 
which is right-angled but not equilateral, a rhombus that 
which is equilateral but not right-angled, and a rhomboid 
that having opposite sides and angles equal to one an- 
other which is neither right-angled nor equilateral. And 
let quadrilateral figures besides these be called trapezia. 

23. Parallel lines are straight-lines which, being in the 
same plane, and being produced to infinity in each direc- 
tion, meet with one another in neither (of these direc- 
tions) . 



t This should really be counted as a postulate, rather than as part of a definition. 



AiTTjfiorca. 

a'. 'Hixrjcrdco duo navxoc; ay]y.eiov era raw arjueTov 
eMelav ypa^rjv dyayelv. 

P'. Kal TC£7iepaa^£v/)v eMelav xaxd xo auve/ec; in 
eCWteiac; expaXeTv. 

y'. Kal Ttavxl xevxpcp xal Biaaxr^axi xuxXov ypdcpecrdai. 

8'. Kal naooLc, xd<; 6pi9a<; ycovtac laac dXXr]Xai<; elvai. 

z. Kal edv sic, 8uo eO'deiat; eui^sTa s^TUTtxouaa xdc; evxoc 
xal Era xa auxd [ispf] ywviac; Suo opiJcov eXdaaovag raoirj, 
expaXXojievac; xag 86o eu-deiac; in aTteipov au^iraTTxeiv, ecp' 
a y.spi] eialv ai x«v Suo 6pi9Gv eXdaaovec;. 



Postulates 

1. Let it have been postulated* to draw a straight-line 
from any point to any point. 

2. And to produce a finite straight-line continuously 
in a straight-line. 

3. And to draw a circle with any center and radius. 

4. And that all right-angles are equal to one another. 

5. And that if a straight-line falling across two (other) 
straight-lines makes internal angles on the same side 
(of itself whose sum is) less than two right-angles, then 
the two (other) straight-lines, being produced to infinity, 
meet on that side (of the original straight-line) that the 
(sum of the internal angles) is less than two right-angles 
(and do not meet on the other side).* 



t The Greek present perfect tense indicates a past action with present significance. Hence, the 3rd-person present perfect imperative 'HiT^cnSa) 
could be translated as "let it be postulated", in the sense "let it stand as postulated", but not "let the postulate be now brought forward". The 
literal translation "let it have been postulated" sounds awkward in English, but more accurately captures the meaning of the Greek, 
t This postulate effectively specifies that we are dealing with the geometry of flat, rather than curved, space. 



Koivat svvoiai. 



Common Notions 



a'. Td xw auxw laa xal dXXrjXou; eaxlv loa. 
P'. Kal edv laoic I'oa TCpoaxcdrj, xd oXa eaxlv laa. 
y'. Kal edv duo ia«v I'oa dcpaipsiS/j, xd xaxaXeiTto^ievd 
eaxiv I'oa. 

8'. Kal xd ecpap[i6£ovxa iiC dXXf]Xa loa dXXrjXou; eaxiv. 
e'. Kal xo oXov xou [Lepouz [leiZov [eaxiv]. 



1. Things equal to the same thing are also equal to 
one another. 

2. And if equal things are added to equal things then 
the wholes are equal. 

3. And if equal things are subtracted from equal things 
then the remainders are equal.* 

4. And things coinciding with one another are equal 
to one another. 

5. And the whole [is] greater than the part. 



t As an obvious extension of C.N.s 2 & 3 — if equal things are added or subtracted from the two sides of an inequality then the inequality remains 



7 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



an inequality of the same type. 



a'. 

'Era xfj? SoOstaiqc; eu-Mac; 7tETtepaa^Evr]<; xpiycovov 
laoTtXeupov auaxi^aaCTdai. 




"Eaxw f] Bo'deToa su'dsla Tt£7tepaa[ievr) r) AB. 
Aei 8r) era xrjg AB su-Mac; xpiywvov [aoTtXsupov 
auaxrjaacrdoa. 

Ksvxpcp jiev xG A 8iaoxr]|jiaxi 8e xG AB xuxXog 
yEypdcpi&M 6 BrA, xal TtdXiv xsvxpcp [Lev xG B Siaaxr^axi 8e 
xG BA xuxXoc; yeypa.^-doi 6 ArE, xal diro xou T orjjieiou, 
xai}' o xsjivouaiv dXXr|Xou<; ol xuxXoi, em xd A, B ar](jieTa 
£Tie^£U)cd«oav eui&eTai di FA, TB. 

Kdi etc el xo A or)(ieTov xevxpov eaxl xou TAB xuxXou, 
lay] eaxlv r) AT xfj AB- TtaXiv, inzi xo B cruie'iov xevxpov 
eaxl xou IAE xuxXou, Xat] eaxlv r] Br xfj BA. eBeix'dr] 8e 
xal f) TA xfj AB iar)' exaxepa dpa xGv IA, TB xfj AB eaxiv 
Tar), xa 8e xG auxG I'oa xal dXXf|Xoic; eaxlv I'aa- xal r) TA dpa 
xfj TB eaxiv lay]- al xpeu; dpa al TA, AB, Br laai dXXf|Xaic; 
siaiv. 

'IaonXsupov dpa eaxl xo ABr xpiycovov. xal auveaxaxai 
era xfjc 8oi9e[ar]c; eMeiac; TtETtepaa^iev/jc; xrjg AB. omep e8ei 
noifjaai. 



t The assumption that the circles do indeed cut one another should be 
that two straight-lines cannot share a common segment. 



Proposition 1 

To construct an equilateral triangle on a given finite 
straight-line. 




Let AB be the given finite straight-line. 

So it is required to construct an equilateral triangle on 
the straight-line AB. 

Let the circle BCD with center A and radius AB have 
been drawn [Post. 3], and again let the circle ACE with 
center B and radius BA have been drawn [Post. 3]. And 
let the straight-lines CA and CB have been joined from 
the point C, where the circles cut one another,* to the 
points A and B (respectively) [Post. 1]. 

And since the point A is the center of the circle CDB, 
AC is equal to AB [Def. 1.15]. Again, since the point 
B is the center of the circle CAE, BC is equal to BA 
[Def. 1.15]. But CA was also shown (to be) equal to AB. 
Thus, CA and CB are each equal to AB. But things equal 
to the same thing are also equal to one another [C.N. 1]. 
Thus, CA is also equal to CB. Thus, the three (straight- 
lines) CA, AB, and BC are equal to one another. 

Thus, the triangle ABC is equilateral, and has been 
constructed on the given finite straight-line AB. (Which 
is) the very thing it was required to do. 

aunted as an additional postulate. There is also an implicit assumption 



P'- 

IIp6<; xG 8oil>£vxi or)\izi<x> xfj So-Oslor) eMteiqc Tarjv eu-delav 
deo-dai. 

'Eaxco xo \iev SotJev ar)|isTov xo A, f) Be 8oif)eTaa eu'dela 
f] Br- 8eT 8fj Ttp6<; xG A arpeta xfj Bo'ddar] eO'deia xfj Br 
iar]v euiMav deo-dai. 

'Erav£eu)c&co yap duo xou A ar)(i£iou era xo B ar)[ieTov 
su-dela f] AB, xal auveaxdxw etc' auxrjc; xpiyovov ia6n:Xeupov 
xo AAB, xal expepXf|aT!}«aav etc' cu-deiac; xau; AA, AB 



Proposition 2" 1 " 

To place a straight-line equal to a given straight-line 
at a given point (as an extremity) . 

Let A be the given point, and BC the given straight- 
line. So it is required to place a straight-line at point A 
equal to the given straight-line BC. 

For let the straight-line AB have been joined from 
point A to point B [Post. 1], and let the equilateral trian- 
gle DAB have been been constructed upon it [Prop. 1.1]. 



8 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



eMelai oci AE, BZ, xal xevxpw |iev iu B Biaaxr^iaxi 8e iu 
Br xuxXoc; yeypdcp-dw 6 THO, xal iraXiv xevxpw xfi A xal 
8iaaxr|uaxi iu AH xuxXoc; yeypdcp-dw 6 HKA. 




E 



'Etc! ouv to B ar^elov xevxpov eaxl xoD rHO, lar\ eaxiv 
f] Br xfj BH. TtdXiv, ETtel xo A arpelov xevxpov eaxl xoO 
HKA xuxXou, iar) eaxiv f] A A xfj AH, Sv f) A A xfj AB Iar] 
eaxiv. Xomf] dpa f) AA XoiTtfj xfj BH eaxiv 'iar). e8e[)cdr] 8s 
xod f] BE xfj BH iar) - exaxepa dpa xfiv AA, Br xfj BH eaxiv 
iar). xd 8s xw auxw laa xal dXXf|Xoi<; eaxiv laa- xal f) AA 
dpa xfj Br eaxiv iar). 

npo<; dpa tw 8oi[)evxi arpeio xo A xfj Bo'deiar] cu-Ma 
xfj Br iar] euiJeTa xelxai f] AA- omp eSei Ttoirjaai. 



t This proposition admits of a number of different cases, depending on 
Euclid invariably only considers one particular case — usually, the most c 



And let the straight-lines AE and BF have been pro- 
duced in a straight-line with DA and DB (respectively) 
[Post. 2] . And let the circle CGH with center B and ra- 
dius BC have been drawn [Post. 3], and again let the cir- 
cle GKL with center D and radius DG have been drawn 
[Post. 3]. 




F 



E 



Therefore, since the point B is the center of (the cir- 
cle) CGH, BC is equal to BG [Def. 1.15]. Again, since 
the point D is the center of the circle GKL, DL is equal 
to DG [Def. 1.15]. And within these, DA is equal to DB. 
Thus, the remainder AL is equal to the remainder BG 
[C.N. 3]. But BC was also shown (to be) equal to BG. 
Thus, AL and BC are each equal to BG. But things equal 
to the same thing are also equal to one another [C.N. 1]. 
Thus, AL is also equal to BC. 

Thus, the straight-line AL, equal to the given straight- 
line BC, has been placed at the given point A. (Which 
is) the very thing it was required to do. 

ie relative positions of the point A and the line BC. In such situations, 
ficult — and leaves the remaining cases as exercises for the reader. 



y'- 

Auo SoOeiafiv eu-deiaiv dviawv duo xrjc; ^ie[£ovo<; xfj 
eXdaaovi l'ar)v eurMav dcpeXelv. 

'Eaxwaav ai Bo-deTaai 8uo euiJeTai dviaoi ai AB, T, 5v 
^ei^wv eaxM f) AB- Be! Bf) aTto xfjc; [ieiZ,o\oc, xfjc; AB xfj 
eXdaaovi xfj T I'arjv eO'delav dcpeXelv. 

KeiadM npbc, x£> A ar)[ie(w xfj T eu'dela iar] f) AA- xal 
xevxpo y.ev xo A 5iaaxf|[jiaxi Be xw AA xuxXoc; yeypdcp-dw 
6 AEZ. 

Kod enel to A ar^elov xevxpov eaxl xoO AEZ xuxXou, 



Proposition 3 

For two given unequal straight-lines, to cut off from 
the greater a straight-line equal to the lesser. 

Let AB and C be the two given unequal straight-lines, 
of which let the greater be AB. So it is required to cut off 
a straight-line equal to the lesser C from the greater AB. 

Let the line AD, equal to the straight-line C, have 
been placed at point A [Prop. 1.2]. And let the circle 
DEF have been drawn with center A and radius AD 
[Post. 3]. 



9 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



Tar) eaxlv f] AE xfj AA- dXXd xal i\ T xfj A A eaxiv for), 
exaxepa apa xov AE, T xfj AA eaxiv Tor)- waxe xai f) AE 
xfj T eaxiv iar). 

r 




Auo apa Bcdeiaaiv euiSeifiv dv[o«v xaiv AB, T duo xfjc; 
(ie[£ovo<; xfj? AB xfj eXdaaovi xfj T iar) dcpfiprjxai f] AE- ojtep 
eSei noifjaai. 

5'. 

Edv 860 xpiywva xa<; 860 nXeupdc; [xalc] Suol rcXeupau; 
Taac; e/r] exaxepav exaxepa xal xf]v y^viav xfj ywvia larjv 
e/rj xf]v Otto xGv ia«v eMeifiv Ttepiexojievrjv, xai T ^ v 
pdaiv xf) pdaei I'arjv e^ei, xal xo xptywvov xw xpiyovo laov 
eaxai, xal ai XoiTtal yoviai xdu; XoiTtau; ywviau; laai eaovxai 
exaxepa exaxepa, Ocp' ac ai laai TtXeupal Gitoxeivouaiv. 



A A 




TiaxM 860 xpiytova xd ABr, AEZ xdc; 860 icXeupac; 
xdc AB, Ar xal<; 8ual TtXeupalg xdu; AE, AZ i'aa<; exovxa 
exaxepav exaxepa xf]v ^.ev AB xfj AE xf)v 8e Ar xfj AZ 
xal y«v(av xf]v utco BAT ywvia xfj Otco EAZ larjv. Xeyco, 
6x1 xal [3daic; f) Br pdaei xrj EZ Tar) eaxiv, xal xo ABr 
xpiyovov xw AEZ xpiywvw I'aov eaxai, xal ai Xoiical ytoviai 
xau; XoiTtalc; ywviau; laai eaovxai exaxepa exaxepa, Ocp'' ac; 
ai laai TtXeupal UTtoxeivouaiv, f) [ie\ utco ABr xrj utco AEZ, 
f] 8e utco ArB xrj utco AZE. 

'Ecpap|jio^o|jievou yap xou ABr xpiywvou era xo AEZ 
xpiycovov xal xi'dejievou xou jiev A ar][ieiou era xo A orj^ielov 



And since point A is the center of circle DEF, AE 
is equal to AD [Def. 1.15]. But, C is also equal to AD. 
Thus, and C are each equal to AD. So is also 
equal to C [C.N. 1]. 

c 




B 



Thus, for two given unequal straight-lines, AB and C, 
the (straight-line) AE, equal to the lesser C, has been cut 
off from the greater AB. (Which is) the very thing it was 
required to do. 

Proposition 4 

If two triangles have two sides equal to two sides, re- 
spectively, and have the angle (s) enclosed by the equal 
straight-lines equal, then they will also have the base 
equal to the base, and the triangle will be equal to the tri- 
angle, and the remaining angles subtended by the equal 
sides will be equal to the corresponding remaining an- 
gles. 



A D 




Let ABC and DEF be two triangles having the two 
sides AB and AC equal to the two sides DE and DF, re- 
spectively. (That is) AB to DE, and AC to DF. And (let) 
the angle BAG (be) equal to the angle EDF. I say that 
the base BC is also equal to the base EF, and triangle 
ABC will be equal to triangle DEF, and the remaining 
angles subtended by the equal sides will be equal to the 
corresponding remaining angles. (That is) ABC to DEF, 
and ACB to DFE. 

For if triangle ABC is applied to triangle DEF,^ the 
point A being placed on the point D, and the straight-line 



10 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



xrjc 8e AB zu'dziac, era xrjv AE, ecpapjioaei xal to B arpeiov 
era to E 8ia to Taiqv elvai x/jv AB xrj AE- ecpap|ioadaT]c; 8f) 
xfjt; AB em TTjv AE ^apjioaei xal f) Ar eMela era xr]v AZ 
6id to ToTjv elvai xr)v utio BAr ywviav xfj utio EAZ' Saxe xal 
to r or)(ieIov era to Z arpelov ecpapjioaei 5ia to larjv TidXiv 
elvai t/)v Ar xfj AZ. dXXa [ir\\> xal xo B era xo E ecprjp^oxei- 
Saxe pdoic; f] Br era pdaiv xrjv EZ ecpapjioaei. el yap xoO 
(iev B era xo E ecpapjioaavxog xoO Be T era xo Z f) Br pdau; 
era xr)v EZ oux ecpapjioaei, 860 eui&eTai /wpiov uepie^ouoiv 
ojiep eaxlv dBuvaxov. ecpapjioaei dpa f] Br pdai<; era x/]v 
EZ xal Tar) auxfj eaxai- Saxe xal oXov xo ABr xpiytovov 
era oXov xo AEZ xpiywvov ecpapjioaei xal I'aov auxcp eaxai, 
xal ai Xomal ycoviai era xdg Xoina? ycoviag ecpapjioaouai xal 
laai auxalc eaovxai, f) ^ev utio ABr xrj utio AEZ fj Be utio 
ArB xrj utio AZE. 

'Eav dpa 860 xpiyiova xac; 8uo TiXeupag [xdig] 660 
TiXeupdic; iaac; evt) exaxepav exaxepa xal xrjv yioviav xrj 
ywvia lorjv exn xrjv utio xfiv Tacov eui5eifiv Tiepie/o|jievr]v, 
xal xr]v pdaiv xf) pdaei Tarjv e<;ei, xal xo xpiywvov xG 
xpiycovco I'aov eaxai, xal ai Xouial ycovlai xalg Xoiraiu; 
ycoviaig laai eaovxai exaxepa exaxepa, ucp' a<; al laai TiXeupal 
UTioxeivouaiv oTiep eBei BeT^ai. 



AB on DE, then the point B will also coincide with E, 
on account of AB being equal to DE. So (because of) 
^4i? coinciding with DE, the straight-line AC will also 
coincide with DF, on account of the angle BAG' being 
equal to EDF. So the point C will also coincide with the 
point F, again on account of AC being equal to DF. But, 
point B certainly also coincided with point E, so that the 
base BC will coincide with the base EF. For if B coin- 
cides with E, and C with F, and the base -BC does not 
coincide with EF, then two straight-lines will encompass 
an area. The very thing is impossible [Post. 1].* Thus, 
the base BC will coincide with EF, and will be equal to 
it [C.N. 4]. So the whole triangle ABC will coincide with 
the whole triangle DEF, and will be equal to it [C.N. 4]. 
And the remaining angles will coincide with the remain- 
ing angles, and will be equal to them [C.N. 4]. (That is) 
ABC to DEF, and ACB to DFE [C.N. 4]. 

Thus, if two triangles have two sides equal to two 
sides, respectively, and have the angle (s) enclosed by the 
equal straight-line equal, then they will also have the base 
equal to the base, and the triangle will be equal to the tri- 
angle, and the remaining angles subtended by the equal 
sides will be equal to the corresponding remaining an- 
gles. (Which is) the very thing it was required to show. 



t The application of one figure to another should be counted as an additional postulate, 
t Since Post. 1 implicitly assumes that the straight-line joining two given points is unique. 



S . 

Tc5v iaoaxeXGv xpiywvwv ai xpog xfj pdaei ywviai iaai 
dXXrjXau; eiaiv, xal TipoaexpXr]i9eia5v xwv lacov eui9ei«v ai 
utio x/]v pdaiv ywviai laai aXXrjXait; eaovxai. 



A 




Tiaxco xpiywvov iaoaxeXec; xo ABr iar]v 'iyov xr]v 
AB TiXeupav xfj AT TiXeupa, xal TipoaexpepXfja'dwaav en 
eCWteiac; xdi<; AB, Ar einMai ai BA, TE- Xeyw, oxi f] [ie\> 
utio ABr ywvia xfj utio ArB Tar) eaxiv, f] Be Otto TBA xrj 
utio BrE. 

EiXrjcp'dco yap era xfjc BA xu/6v ar^ieiov xo Z, xal 
dcprjpria'dco arco xfj? [iei£ovo<; xfjc; AE xfj eXdaaovi xfj AZ 



Proposition 5 

For isosceles triangles, the angles at the base are equal 
to one another, and if the equal sides are produced then 
the angles under the base will be equal to one another. 

A 




D E 



Let ABC be an isosceles triangle having the side AB 
equal to the side AC, and let the straight-lines BD and 
CE have been produced in a straight-line with AB and 
AC (respectively) [Post. 2]. I say that the angle ABC is 
equal to ACB, and (angle) CBD to BCE. 

For let the point F have been taken at random on BD, 
and let AG have been cut off from the greater AE, equal 



11 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



Tar] f] AH, xai £Tce£eu)(iL>Gjaav ai Zr, HB sMeiai. 

'EtceI ouv iar] eaxlv f] [iev AZ xfj AH f] 8e AB xfj Ar, 
8uo 8f| al ZA, Ar Sua! xdic. HA, AB laai elolv exaxepa 
sxaxspor xal ywviav xoivrjv Tcepii)(ouai xf)v utco ZAH' pdaic, 
apa f) Zr pdaei xfj HB Tar) eaxiv, xal xo AZr xpiyovov iu 
AHB xpiyovo Taov eaxai, xal ai Xomal ycoviai toac. XomaTc. 
ycoviaic, I'oai eaovxai exaxepa exaxepa, ucp' ac, ai Taai TcXeupal 
UTtoxeivouaiv, r] ^ev utco ATZ xfj utco ABH, f) 8e utco AZr 
xfj utco AHB. xal CTcel 6Xt] r] AZ oXy] xfj AH eaxiv Tar], Gv 
f) AB xfj Ar eaxiv Tar], XoiTcf] apa f] BZ Xomfj xfj TH eaxiv 
larj. eBeix^r] Se xal f] Zr xfj HB for) - Suo 8f] ai BZ, Zr 8ual 
xdic. TH, HB Taai eiaiv exaxepa exaxepa- xal yiovia f] utco 
BZr yiovia xrj utco THB lor), xal pdaic auxcov xoivf] f) BT- 
xal xo BZr apa xpiywvov x« THB xpiycbvcp Taov eaxai, xal 
ai XoiTtal ywviai xdic XoiTcdic ycoviaic Taai eaovxai exaxepa 
exaxepa, ucp' ac ai i'oai TcXeupal UTioxeivouaiv I'or) dpa eaxlv 
f) ^iev utco ZBr xfj utto HTB f] Se utco BrZ xfj Gtco TBH. 
etxeI ouv 6Xt) f] utco ABH ytovia oXr) xfj utco ArZ ytovia 
£8ei)cdr) iar), cov f] utco TBH xfj utco BrZ lar], Xoiti/) apa f] 
utco ABr XoiTxfj xfj utco ArB screw iar) - xai eiai Tcpoc xfj 
pdaei xou ABr xpiytovou. eSei/'dr] 8s xal f] utco ZBr xfj 
utco HrB Tar)- xai eiaiv utco xf)v pdaiv. 

Tcov dpa iaoaxeXGv xpiytovtov ai xpoc xfj pdaei ycoviai 
Taai dXXfjXaic eiaiv, xai TcpoaexpXrji&eiaGv xfiv Tacov euifteicov 
ai utco xf]v pdaiv ycoviai Taai dXXrjXaic eaovxai- OTcep eSei 
8eT^ai. 



9 . 

'Edv xpiycovou ai 8uo ycoviai Taai dXXf]Xaic coaiv, xal 
ai utco xdc Taac ycoviac UTCoxeivouaai TcXeupal Taai dXXr]Xaic 
eaovxai. 



A 




'Eaxco xpiycovov xo ABr Tarjv e)(ov xrjv utco ABr ycoviav 
xfj utco ArB ycovia- Xeyo, oxi xal TcXeupd f] AB TcXeupa xfj 
Ar eaxiv iar). 



to the lesser AF [Prop. 1.3]. Also, let the straight-lines 
FC and GB have been joined [Post. 1]. 

In fact, since AF is equal to AG, and AB to AC, 
the two (straight-lines) FA, AC are equal to the two 
(straight-lines) GA, AB, respectively. They also encom- 
pass a common angle, FAG. Thus, the base FC is equal 
to the base GB, and the triangle AFC will be equal to the 
triangle AGB, and the remaining angles subtendend by 
the equal sides will be equal to the corresponding remain- 
ing angles [Prop. 1.4]. (That is) ACF to ABG, and AFC 
to AGB. And since the whole of AF is equal to the whole 
of AG, within which AB is equal to AC, the remainder 
BF is thus equal to the remainder CG [C.N. 3]. But FC 
was also shown (to be) equal to GB. So the two (straight- 
lines) BF, FC are equal to the two (straight-lines) CG, 
GB, respectively, and the angle BFC (is) equal to the 
angle CGB, and the base BC is common to them. Thus, 
the triangle BFC will be equal to the triangle CGB, and 
the remaining angles subtended by the equal sides will be 
equal to the corresponding remaining angles [Prop. 1.4]. 
Thus, FBC is equal to CGB, and BCF to CBG. There- 
fore, since the whole angle ABG was shown (to be) equal 
to the whole angle ACF, within which CBG is equal to 
BCF, the remainder ABC is thus equal to the remainder 
ACB [C.N. 3]. And they are at the base of triangle ABC. 
And FBC was also shown (to be) equal to GCB. And 
they are under the base. 

Thus, for isosceles triangles, the angles at the base are 
equal to one another, and if the equal sides are produced 
then the angles under the base will be equal to one an- 
other. (Which is) the very thing it was required to show. 

Proposition 6 

If a triangle has two angles equal to one another then 
the sides subtending the equal angles will also be equal 
to one another. 

A 




Let ABC be a triangle having the angle ABC equal 
to the angle ACB. I say that side AB is also equal to side 

AC. 



12 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



El yap aviao<; eaxiv f] AB xfj Ar, rj exepa aOxCSv ^.el^tov 
eaxlv. eaxo (jlel^cov f\ AB, xal dcprjpfia'do dra 5 ) xfj? ^el^ovoc; 
xfj? AB xfj eXdxxovi xfj Ar lot) f) AB, xal eTie£eu)(i&Gj ^ AT. 

'End ouv Xat] eaxlv f] AB xrj Ar xoivf] 8s r) Br, 8uo 8f] 
at AB, Br 8uo xau; Ar, TB I'aai elalv exaxepa exaxepa, xal 
ycovla f] Otto ABr ycovla xfj utio ArB eaxiv I'ar)- pdaic; dpa #j 
Ar pdaei xrj AB lor] eaxlv, xal xo ABr xplycovov xfi ArB 
xpiycovco laov eaxai, xo eXaaaov xfi [isl^ovi - oTiep dxoTtov 
oux dpa dviaoc; eaxiv f) AB xfj Ar - I'ar) dpa. 

'Edv dpa xpiycovou al 6uo ycovlai laai dXXrjXau; Saiv, xal 
at Otio xdc; laac; ycovlac; UTtoxelvouaai TtXeupal laai dXXrjXau; 
eaovxau OTiep eSei Set^ai. 



For if AB is unequal to AC then one of them is 
greater. Let AB be greater. And let DB, equal to 
the lesser AC, have been cut off from the greater AB 
[Prop. 1.3]. And let DC have been joined [Post. 1]. 

Therefore, since DB is equal to AC, and BC (is) com- 
mon, the two sides DB, BC are equal to the two sides 
AC, CB, respectively, and the angle DBC is equal to the 
angle ACB. Thus, the base DC is equal to the base AB, 
and the triangle DBC will be equal to the triangle ACB 
[Prop. 1.4], the lesser to the greater. The very notion (is) 
absurd [C.N. 5]. Thus, AB is not unequal to AC. Thus, 
(it is) equals 

Thus, if a triangle has two angles equal to one another 
then the sides subtending the equal angles will also be 
equal to one another. (Which is) the very thing it was 
required to show. 



t Here, use is made of the previously unmentioned common notion that if two quantities are not unequal then they must be equal. Later on, use 
is made of the closely related common notion that if two quantities are not greater than or less than one another, respectively, then they must be 
equal to one another. 



'Era xrjc; auxfjc eui!)ela<; 8uo xau; auxale; eO'delau; aXXai 
8uo euiJkTaL iaai exaxepa exaxepa ou auaxai3f]aovxai Ttpoc; 
aXXcp xal aXXco arpeltp era xd auxd fiepr] xd auxd Tiepaxa 
exouaai xau; z\ dp/fjc; euiitelau;. 




El yap Suvaxov, era iff auxfji; eMelac; iff AB 8uo xdic; 
auxau; eCcdelau; xau; AT, TB dXXai 8uo eu'deTai at AA, AB 
laai exaxepa exaxepa auveaxdxoaav Tipoc; aXXcp xal dXXcp 
ay][ie'icd i& is r xal A era xd auxd ^eprj xd auxd Tiepaxa 
e/ouaai, coaxe lar\v eTvai xf)v [Lev IA xfj AA xo auxo rcepac; 
e/ouaav auxfj xo A, xf)v 8e TB xrj AB xo auxo Tiepac; exou- 
aav auxfj xo B, xal eTieCeuy^dw ^ TA. 

'Erad ouv Xar\ eaxlv r\ AT xfj AA, Tar) tail xal ycovla f] 
utio ATA xfj utio AAT- ^tel^cov dpa f] utio AAT xrjc; utio 
ATB- tioXXco dpa f] utio TAB ueli^cov eaxl xfj? utio ATB. 
TcdXiv STtel Tar] eaxlv f) TB xfj AB, Tar] eaxl xal ycovla f) 
utio TAB ycovla xfj utio ATB. eSelx'dr) 8e auxrjc xal TtoXXcp 
HelC^cov oTiep eaxlv d8uvaxov. 

Oux dpa era xrjc auxfjc eui9elac. 86o xau; auxdu; eu'delau; 



Proposition 7 

On the same straight-line, two other straight-lines 
equal, respectively, to two (given) straight-lines (which 
meet) cannot be constructed (meeting) at a different 
point on the same side (of the straight-line), but having 
the same ends as the given straight-lines. 




For, if possible, let the two straight-lines AC, CB, 
equal to two other straight-lines AD, DB, respectively, 
have been constructed on the same straight-line AB, 
meeting at different points, C and D, on the same side 
(of AB), and having the same ends (on AB). So CA is 
equal to DA, having the same end A as it, and CB is 
equal to DB, having the same end B as it. And let CD 
have been joined [Post. 1]. 

Therefore, since AC is equal to AD, the angle ACD 
is also equal to angle ADC [Prop. 1.5]. Thus, ADC (is) 
greater than DCB [C.N. 5]. Thus, CDB is much greater 
than DCB [C.N. 5]. Again, since CB is equal to DB, the 
angle CDB is also equal to angle DCB [Prop. 1.5]. But 
it was shown that the former (angle) is also much greater 



13 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



dXXai 8uo eui5eTauaai exaxepa exaxepa auaTocdfjaovToa Ttpo? 
dXXcp xal dXXw arj^eio era xd aOxd [Jiepr] xd auxd Ttepaxa 
e/ouaai xal? 15 dp/f)? eOiiteiai?- oitep e8ei BeT^ai. 



'Edv 8uo xpiycova xd? 860 iiXeupd? [xal?] 860 TiXeupal? 
laa? l)(T] exaxepav exaxepa, zyr\ 8e xal xrjv pdaiv xfj pdaei 
larjv, xal xrjv ywviav xfj yiovia I'arjv eijei xf]v unb xwv Tacov 
eMeiwv Ttepiexojiev/jv. 




15axw 860 xpiytova xd ABr, AEZ xd? Buo TtXeupd? 
xd? AB, Ar xai? 660 nXeupaT? xal? AE, AZ !aa? exovxa 
exaxepav exaxepa, xrjv [lev AB xfj AE xf]v 8s Ar xfj AZ- 
8e xal pdaiv xrjv Br pdaei xfj EZ l'ar]v Xey«, oxi xal 
ycovia f) Ono BAr ywvia xfj 6716 EAZ eaxiv iar). 

'E(fctp\ioZo[Levo\j yap xou ABr xpiywvou era xo AEZ 
xpiyovov xal xide^tevou xou \±ev B af][ie'iou era xo E arpeTov 
xfj? Be Br euT!)e[a? era x/]v EZ ecpapjioaei xal xo T arpelov 
excl xo Z Bid xo iar]v elvai xf]v Br xfj EZ- ecpapjioadar]? 8rj 
xfj? Br Ira xrjv EZ ecpap^toaouai xal al BA, TA era xd? EA, 
AZ. ei yap pdai? [lev f\ BT era pdaiv xf)v EZ ecpap^toaei, ai 
8e BA, Ar TiXeupal era xd? EA, AZ oux ecpapjioaouaiv 
dXXd TtapaXXd^ouaiv cb? ai EH, HZ, auaxa'driaovxai era xfj? 
auxfj? euiJeia? 860 xaT? auxaT? eu-deiai? dXXai 860 eu-delai 
laai exaxepa exaxepa rcpo? aXXo xal dXXto ar^eiw era xd 
auxa [ispf] xd auxa Ttepaxa s/ouaai. ou auviaxavxai Be- 
oux dpa ecpap^oi^o^evr)? xfj? Br pdaew? era xrjv EZ pdaiv 
oux ecpapuoaouai xal ai BA, Ar TiXeupal era xd? EA, AZ. 
ecpapjioaouaiv dpa- waxe xal ywvia f] Otco BAr era ywviav 
xf|V Oko EAZ ecpap^toaei xal iar) auxfj eaxai. 

'Edv dpa Buo xpiywva xd? Suo iiXeupd? [xal?] 860 
TiXeupal? I'aa? e^T] exaxepav exaxepa xal xrjv pdaiv xfj pdaei 
lar]v e^T), xal xf]v ytoviav xfj yovia ia/]v e^ei xf)v utco xwv 
lawv eu-deiwv Ttepiexo^evrjv oTtep eBei SeT^ai. 



(than the latter). The very thing is impossible. 

Thus, on the same straight-line, two other straight- 
lines equal, respectively, to two (given) straight-lines 
(which meet) cannot be constructed (meeting) at a dif- 
ferent point on the same side (of the straight-line), but 
having the same ends as the given straight-lines. (Which 
is) the very thing it was required to show. 

Proposition 8 

If two triangles have two sides equal to two sides, re- 
spectively, and also have the base equal to the base, then 
they will also have equal the angles encompassed by the 
equal straight-lines. 




Let ABC and DEF be two triangles having the two 
sides AB and AC equal to the two sides DE and DF, 
respectively. (That is) AB to DE, and AC to DF. Let 
them also have the base BC equal to the base EF. I say 
that the angle BAC is also equal to the angle EDF. 

For if triangle ABC is applied to triangle DEF, the 
point B being placed on point E, and the straight-line 
BC on EF, then point C will also coincide with F, on 
account of BC being equal to EF. So (because of) BC 
coinciding with EF, (the sides) BA and CA will also co- 
incide with ED and DF (respectively) . For if base BC 
coincides with base EF, but the sides AB and AC do not 
coincide with ED and DF (respectively), but miss like 
EG and GF (in the above figure), then we will have con- 
structed upon the same straight-line, two other straight- 
lines equal, respectively, to two (given) straight-lines, 
and (meeting) at a different point on the same side (of 
the straight-line), but having the same ends. But (such 
straight-lines) cannot be constructed [Prop. 1.7]. Thus, 
the base BC being applied to the base EF, the sides BA 
and AC cannot not coincide with ED and DF (respec- 
tively). Thus, they will coincide. So the angle BAC will 
also coincide with angle EDF, and will be equal to it 
[C.N. 4]. 

Thus, if two triangles have two sides equal to two 
side, respectively, and have the base equal to the base, 



14 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



then they will also have equal the angles encompassed 
by the equal straight-lines. (Which is) the very thing it 
was required to show. 



0'. 

Tf|v Softeiaav yioviav eMuypa^ov 5(xa xe^elv. 

A 




'Eaxco f] So'deiaa y«v(a eO'duypa^oc; f] hub BAr. 8eT 
8r) auxfjv Si^a xejieTv. 

EtXfiqydw iiti xfjc AB xu^ov GTj^ielov to A, xal dcpr)pf|C7d« 
dno Tfjc; Ar xfj AA Tar) rj AE, xal eite^eux^M f) AE, xal 
auveaxaxo) em xfjc AE xpiywvov taoTtXeupov xo AEZ, xal 
eTisCeux'dw f] AZ- Xeyto, 6x1 f\ O716 BAr ycovia Bl^a xexurjxai 
O716 xfjc AZ eu'ddac. 

Tkel yap for] eaxlv f] AA xfj AE, xoivf) 8e f\ AZ, 860 8f) 
at AA, AZ Sual xau; EA, AZ foai eiatv exaxepa exaxepa. 
xal pdai<; f] AZ pdaei xfj EZ far) eaxtv ywvia apa r\ bub 
AAZ ywvia xfj O716 EAZ for] eaxtv. 

TL apa 5odefoa yovta eiWJuypa^ot; fj O716 BAr 8txa 
xex^.r)xai utco xfjc AZ eu'detac;- onep e8ei uoifjaai. 



Proposition 9 
To cut a given rectilinear angle in half. 



A 




Let BAC be the given rectilinear angle. So it is re- 
quired to cut it in half. 

Let the point D have been taken at random on AB, 
and let AE, equal to AD, have been cut off from AC 
[Prop. 1.3], and let DE have been joined. And let the 
equilateral triangle DEF have been constructed upon 
DE [Prop. 1.1], and let AF have been joined. I say that 
the angle BAC has been cut in half by the straight-line 
AF. 

For since AD is equal to AE, and AF is common, 
the two (straight-lines) DA, AF are equal to the two 
(straight-lines) EA, AF, respectively. And the base DF 
is equal to the base EF. Thus, angle DAF is equal to 
angle EAF [Prop. 1.8]. 

Thus, the given rectilinear angle BAC has been cut in 
half by the straight-line AF. (Which is) the very thing it 
was required to do. 



l . 

Tf]v Sodefoav eu^ETav Tcenepaajievrjv 8i);a xejieTv. 

Tiaxa> f] Sodefoa eO'dela Keiiepaa^evr) f) AB- 8eT 8f] xf]v 
AB euiMav TteTtepaauevrjv Btya xe^ielv. 

Suveaxdxo en auxfjc; xptywvov taoTtXeupov xo ABT, xal 
xex^ir|<yda> f) Gtco ATB yovta 8i)(a xfj EA euifteta- Xeyw, oxi 
f] AB eu'dela 8txa xex^rjxai xaxd xo A ar][ieTov. 

Tkel yap Tar) eaxlv f\ AT xfj TB, xoivf] 8s f] TA, 860 8f] 
at AT, TA 860 xau; BT, TA foai eiatv exaxepa exaxepa- xal 
ycovta f] O716 ATA ywvta xfj O716 BTA for] eaxtv pdau; apa 



Proposition 10 

To cut a given finite straight-line in half. 

Let AB be the given finite straight-line. So it is re- 
quired to cut the finite straight-line AB in half. 

Let the equilateral triangle ABC have been con- 
structed upon (AB) [Prop. 1.1], and let the angle ACB 
have been cut in half by the straight-line CD [Prop. 1.9]. 
I say that the straight-line AB has been cut in half at 
point D. 

For since AC is equal to CB, and CD (is) common, 



15 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



f) AA (3aa£i xfj BA far) eaxiv. 




'H apa Scdelaa euifteTa TiETtepaa^evr] f\ AB 8[/a xexurjxai 
xaxa to A- oracp eSei rcoifjaai. 



the two (straight-lines) AC, CD are equal to the two 
(straight-lines) BC, CD, respectively. And the angle 
ACD is equal to the angle BCD. Thus, the base AD 
is equal to the base BD [Prop. 1.4]. 

c 




Thus, the given finite straight-line AB has been cut 
in half at (point) D. (Which is) the very thing it was 
required to do. 



ia'. 

Tfj 8o$£[ar) sMeia dno tou npbc, auxrj BoiSevxoc; ar^dou 
npbq 6pM<; ywviai; eMeiav ypa^rjv dyayeiv. 

z 




ATE 

'Eaxco f] ^lev BoiSefaa euiiteTa f) AB to 8e bo-&e\ af][ieiov 
in' auxrjg to T- 8eT Srj arco tou T ar][ieio\j xfj AB cu-Ma 
npoc opMg ytoviac; sMeTav ypaji|if]v dyayeiv. 

EiX^-dtd era xrjc; Ar xu)(6v a/jjislov to A, xal xeia-du 
if) IA Tor) f) TE, xal auvsaxdxto era xrjc AE xpiywvov 
iooTiXeupov xo ZAE, xal inzt.zux'Qid f] ZE Xeyto, oxi xfj 
SoiEteiar) EUifrda xfj AB duo tou upoc auxfj 8oi9£vto<; ar)[idou 
tou r Tipoc; op-dac; yioviac; euiSsIa ypajijif) rjxxai f\ ZT. 

'End yap far) eaxlv f) AT xfj TE, xoivf) 8e f] TZ, 8uo 
8f) ai AT, TZ 8ual xal<; ET, TZ faai dalv exaxspa sxaxepa- 
xal pdaig T) AZ pdaeri xfj ZE lor) eaxiv ywvia apa f) uko 
ATZ ywvta xfj (mo ETZ far] eaxtv xai efaiv £cpe^fj<;. oxav 
8e eu'dda Ik' eiWteTav axaiftefaa xa<; ecpec;fj<; ywv[a<; faa<; 
dXXf)Xaic rcoifj, op'df] exaxspa xwv fawv ywviwv eaxiv op'df) 
apa eaxlv exaxepa xwv utio ATZ, ZTE. 



Proposition 11 

To draw a straight-line at right-angles to a given 
straight-line from a given point on it. 

F 




D C E 

Let AB be the given straight-line, and C the given 
point on it. So it is required to draw a straight-line from 
the point C at right-angles to the straight-line AB. 

Let the point D be have been taken at random on AC, 
and let CE be made equal to CD [Prop. 1.3], and let the 
equilateral triangle FDE have been constructed on DE 
[Prop. 1.1], and let EC have been joined. I say that the 
straight-line EC has been drawn at right-angles to the 
given straight-line AB from the given point C on it. 

For since DC is equal to CE, and CF is common, 
the two (straight-lines) DC, CF are equal to the two 
(straight-lines), EC, CF, respectively. And the base DF 
is equal to the base FE. Thus, the angle DCF is equal 
to the angle ECF [Prop. 1.8], and they are adjacent. 
But when a straight-line stood on a(nother) straight-line 



1G 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



Tfj apa BoiDeior] euiSeia xfj AB and xoO Ttp6<; auxrj 
BcedevTcx; ar)^e(ou tou T Tcpoc; dpfiuc, ywviac; eO'deTa ypa^r) 
rjxxai f] TZ- oitep eBei noirjaai. 



10'. 

'Era x/]v 8oi!)eTaav eu'deuxv aneipov &tco xoO So-devxoc 
arj^eiou, 8 [ir\ eaxiv etc' auxfj<;, xdiJexov euiMav ypa^ir]v 
dyayeTv. 

z 




"Eaxo f] ^tev So'de'iaa eu'de'ia aiceipcx; f] AB xo 8e 8oi9ev 
ar)^eTov, 8 ^ eaxiv etc' auxfj<;, xo E 8eT Br] era xrjv Bo-deTaav 
eMelav dneipov xr]v AB duo xoO Scdevxot; ar)|J.e[ou xoO T, 
8 \ir\ eaxiv etc' auxfjc;, xdcdexov eu-delav ypa^rjv dyayeTv. 

EiXr]cpdco yap tiii xd exepa jiepr] xfj<; AB eu^eiac; xu^ov 
arjjieTov xo A, xdi xevxpw [lev xai T 8iaaxr]jjiaxi 8e xa> EA 
xuxXoc; yeypdcpiJw 6 EZH, xdi xex|ir|ada) f] EH eui5ela 8i/a 
xaxd xo 6, xdi C7te£eu)(iL)Maav di TH, TO, TE euiiteTai- 
Xeya>, oxi era xr)v BcdeTaav euaDeTav aracipov xrjv AB duo 
xou Bo-devxoc; at]\ieiou xou T, 3 \Lr\ eaxiv sit' auxrjc;, xd-dexoc; 
rjxxai f] TO. 

'Eracl ydp Xar\ eaxiv f) HO xfj 8E, xoivr] Be f) Or, Buo 
8/) di H0, OT Buo xau; EO, Or Taai eialv exaxepa exaxepa- 
xdi pdaic; f) TH [3daei xfj TE eaxiv lot]- yiovia apa f) Otto 
T0H ya>v[a xrj Otto E0r eaxiv Tar), xai eiaiv ecpe^fjg. oxav 
Be eO'deTa en' eui5eTav axa-delaa xdc; ecpe^rjc ywviac; Taag 
dXXrjXaic; Ttoifj, 6p$r) exaxepa xwv Tacov ycoviwv eaxiv, xai 
f) ecpeaxrjxula eui5ela xd-dexog xaXelxai ecp' fjv ecpeaxrjxev. 

'Era xrjv Bo-deTaav apa eu-delav aracipov xrjv AB duo xou 
Boi&evxoc; ar\[ieiou xou T, o ^fj eaxiv in auxfjc, xd-dexoc; 
rjxxai f] TO- oitep eBei icoifjaai. 



makes the adjacent angles equal to one another, each of 
the equal angles is a right-angle [Def. 1.10]. Thus, each 
of the (angles) DCF and FCE is a right-angle. 

Thus, the straight-line CF has been drawn at right- 
angles to the given straight-line AB from the given point 
C on it. (Which is) the very thing it was required to do. 

Proposition 12 

To draw a straight-line perpendicular to a given infi- 
nite straight-line from a given point which is not on it. 



F 




D 

Let AB be the given infinite straight-line and C the 
given point, which is not on (AB). So it is required to 
draw a straight-line perpendicular to the given infinite 
straight-line AB from the given point C, which is not on 
{AB). 

For let point D have been taken at random on the 
other side (to C) of the straight-line AB, and let the 
circle EFG have been drawn with center C and radius 
CD [Post. 3], and let the straight-line EG have been cut 
in half at (point) H [Prop. 1.10], and let the straight- 
lines CG, CH, and CE have been joined. I say that the 
(straight-line) CH has been drawn perpendicular to the 
given infinite straight-line AB from the given point C, 
which is not on (AB) . 

For since GH is equal to HE, and HC (is) common, 
the two (straight-lines) GH, HC are equal to the two 
(straight-lines) EH, HC, respectively, and the base CG 
is equal to the base CE. Thus, the angle CHG is equal 
to the angle EHC [Prop. 1.8], and they are adjacent. 
But when a straight-line stood on a(nother) straight-line 
makes the adjacent angles equal to one another, each of 
the equal angles is a right-angle, and the former straight- 
line is called a perpendicular to that upon which it stands 
[Def. 1.10]. 

Thus, the (straight-line) CH has been drawn perpen- 
dicular to the given infinite straight-line AB from the 



17 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



'Eav su'dsla in eMsiav aiadzica. ywviac; Tioifj, fjxoi 8uo 
dp'&a.z fj 8uatv op-ddic; Taa<; TioirjaEi. 




a b r 

EuiJeTa yap xi? f] AB etc' eu-deiav xrjv TA axaiMaa 
yov(a<; tcoieixco xdc utio TBA, ABA- Xsy", oxi at utio TBA, 
ABA ycovtai rjxoi 860 op'dai Eiaiv fj Sualv opiDau; I'aai. 

EE [ie\ ouvTar) saxlv f] utio TBA xfj utio ABA, Suo 6p$a[ 
Eiaiv. si 8s ou, f\x$(i> aTio xou B arj^isiou xfj TA [sMsia] npbc, 
opiDat; f) BE' at apa utio TBE, EBA 860 op'dai Eiaiv- xal 
etieI f] utio TBE Sua! xdic utio TBA, ABE I'ar) saxlv, xoivr] 
Ttpoaxsia'dw f) utio EBA- at apa utio TBE, EBA xpial xau; 
utio TBA, ABE, EBA I'aai Eiaiv. TidXiv, etcsi f) utio ABA 
Sua! xau; utio ABE, EBA Tar) saxlv, xoivr) Tipoaxsla'dw f) 
utio ABE at apa utio ABA, ABr xpial xau; utio ABE, EBA, 
ABr Taai siaiv. sSslxi&rjaav 8s xal at utio TBE, EBA xpial 
xdic; auxaii; Taai- xa 8s xo auxo Taa xal dXXr]Xou; saxlv Taa- 
xal at utio TBE, EBA dpa xau; utio ABA, ABr Taai siaiv- 
dXXd at utio TBE, EBA Suo op'dai eiaiv xal at utio ABA, 
ABr dpa Sualv opiJau; Taai Eiaiv. 

'Eav apa su-dsla in suiiteTav axaiSsTaa ywvlac; Tioifj, f)xoi 
Suo op'ddi; fj 6ualv opiSdic; Taac; TioirjaEi- oTisp sSsi 5sTc;ai. 



given point C, which is not on ( AB) . (Which is) the very 
thing it was required to do. 

Proposition 13 

If a straight-line stood on a(nother) straight-line 
makes angles, it will certainly either make two right- 
angles, or (angles whose sum is) equal to two right- 
angles. 




D B C 

For let some straight-line AB stood on the straight- 
line CD make the angles CBA and ABD. I say that 
the angles CBA and ABD are certainly either two right- 
angles, or (have a sum) equal to two right-angles. 

In fact, if CBA is equal to ABD then they are two 
right-angles [Def. 1.10]. But, if not, let BE have been 
drawn from the point B at right-angles to [the straight- 
line] CD [Prop. 1.11]. Thus, CBE and EBD are two 
right-angles. And since CBE is equal to the two (an- 
gles) CBA and ABE, let EBD have been added to both. 
Thus, the (sum of the angles) CBE and EBD is equal to 
the (sum of the) three (angles) CBA, ABE, and EBD 
[C.N. 2]. Again, since DBA is equal to the two (an- 
gles) DBE and EBA, let ABC have been added to both. 
Thus, the (sum of the angles) DBA and ABC is equal to 
the (sum of the) three (angles) DBE, EBA, and ABC 
[C.N. 2]. But (the sum of) CBE and EBD was also 
shown (to be) equal to the (sum of the) same three (an- 
gles) . And things equal to the same thing are also equal 
to one another [C.N. 1]. Therefore, (the sum of) CBE 
and EBD is also equal to (the sum of) DBA and ABC. 
But, (the sum of) CBE and EBD is two right-angles. 
Thus, (the sum of) ABD and ABC is also equal to two 
right-angles. 

Thus, if a straight-line stood on a(nother) straight- 
line makes angles, it will certainly either make two right- 
angles, or (angles whose sum is) equal to two right- 
angles. (Which is) the very thing it was required to show. 



18 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



18'. 

'Eav Tcpoc; tlvi sMsia xal xG Tcpoc auxfj arista 860 
su-dslai [L7] era xa auxa [iepf] xsi^isvai xag scpsi;fjc; ycovia? 
8uolv opiSalg iaa<; rcoiScnv, en' sui)siac; saovxai dXXfjXau; ai 
su-dslai. 



A E 




r b A 

LTpoc; yap xivi su-dsia xfj AB xai xco upog auxfj arjjjieLtp 
x£> B 860 su-dslai ai Br, BA [if] era xa auxa (icpr) xeijievai 
xa<; ecpe^fj<; y«via<; xa<; utco ABr, ABA 860 op-dau; laac; 
Tioieixwaav Xsy«, oxi etc' euiSetai; eaxi xfj TB f] BA. 

El yap sqxi xfj Br etc' su-Ssiac; f) BA, eax« xfj TB 
etc' su-dsia^ f] BE. 

'Etcei ouv su-dsla f) AB etc' su-dslav xf]v TBE scp£axr]xsv, 
ai apa utco ABr, ABE ytoviai Suo op-dau; laai siaiv siai 8s 
xai ai utco ABr, ABA 860 opiSaTi; foai - ai apa utco TBA, 
ABE xaT? utco TBA, ABA laai eiaiv. xoivf] dcpr)pf|a , dw f) 
utco TBA- Xoitct) apa f] utco ABE XoiTcfj xfj utco ABA saxiv 
lay], i] sXaaawv xfj [is[£ovr oTcsp saxiv dSuvaxov. oux dpa 
etc' sMsiac; saxiv f] BE xfj TB. ojioiok 8f] 8sic;o^£v, oxl ouSs 
dXXr) xu; TcXfjv xfjc; BA- etc' su-dsiac; dpa saxiv f) TB xfj BA. 

'Eav apa Tcpog xivi sui5sia xai cfi Tcpoc; auxfj arista 
860 su^siai [iv] sra auxa [tept] xsipisvai xdg scps^fjg yioviac; 
Suoiv op-ddig Taac; Tcouoaiv, etc' su-dsiac saovxai dXXfjXaic; ai 
su-dslai- oTcsp sSsi Ssl^ai. 



is'. 

'Eav 860 sO'ds'iai xsjivwaiv dXXf|Xac;, xac; xaxa xopucpfjv 
ycovia<; laag dXXfjXau; tcoiouoiv. 



Proposition 14 

If two straight-lines, not lying on the same side, make 
adjacent angles (whose sum is) equal to two right-angles 
with some straight-line, at a point on it, then the two 
straight-lines will be straight-on (with respect) to one an- 
other. 



A E 




C B D 



For let two straight-lines BC and BD, not lying on the 
same side, make adjacent angles ABC and ABD (whose 
sum is) equal to two right-angles with some straight-line 
AB, at the point B on it. I say that BD is straight-on with 
respect to CB. 

For if BD is not straight-on to BC then let BE be 
straight-on to CB. 

Therefore, since the straight-line AB stands on the 
straight-line CBE, the (sum of the) angles ABC and 
ABE is thus equal to two right-angles [Prop. 1.13]. But 
(the sum of) ABC and ABD is also equal to two right- 
angles. Thus, (the sum of angles) CBA and ABE is equal 
to (the sum of angles) CBA and ABD [C.N. 1]. Let (an- 
gle) CBA have been subtracted from both. Thus, the re- 
mainder ABE is equal to the remainder ABD [C.N. 3], 
the lesser to the greater. The very thing is impossible. 
Thus, BE is not straight-on with respect to CB. Simi- 
larly, we can show that neither (is) any other (straight- 
line) than BD. Thus, CB is straight-on with respect to 
BD. 

Thus, if two straight-lines, not lying on the same side, 
make adjacent angles (whose sum is) equal to two right- 
angles with some straight-line, at a point on it, then the 
two straight-lines will be straight-on (with respect) to 
one another. (Which is) the very thing it was required 
to show. 

Proposition 15 

If two straight-lines cut one another then they make 
the vertically opposite angles equal to one another. 



19 



ETOIXEIftN a'. 



ELEMENTS BOOK 1 



Auo yap eu"delai at AB, TA xejivexwaav dXXf|Xac; xaxa 
to E arjueiov Xeyco, 8x1 iar) eaxlv rj (iev utio AEr ycovta xfj 
utto AEB, f) 8e utto TEB xfj Otto AEA. 

A 




B 

Tkel yap eO-dsTa f\ AE eu^slav xfjv TA ecpeaxrjxe 
y«v(a<; Tioiouaa xdc; Otto TEA, AEA, at apa utio TEA, AEA 
yovtai 8ualv dp'&aa.q I'aai eiaiv. TtdXiv, ercel eui5ela f] AE eit' 
eMelav xf]v AB ecpeaxrjxe yioviag noiouaa xdg utio AEA, 
AEB, at apa utio AEA, AEB ycoviai 8ualv opiate; Xaai etatv. 
e8d)edr)o-av 8e xal at Otto TEA, AEA Buotv opiate; Taai- at 
apa utio TEA, AEA xau; utio AEA, AEB laai slow, xoivrj 
d(pr)pf|adco f) utio AEA- Xouif] apa f) Otto TEA Xoutfj xfj utto 
BEA far) eaxtv o^totwc; 8f] Beix^rjaexai, oxi xal at Otto TEB, 
AEA laai eiatv. 

'Eav apa 8uo eui9eTai xe^tvwaiv dXXf|Xa<;, xa<; xaxa xo- 
pucprjv ywvtag laac, dXXrjXaii; ttoiouoiv orcep e8ei SeT^oti. 



If'. 

navxog xpiycivou (iiac xfiv TtXeupwv TtpoaexpXr] , Mar)<; 
f) exxoc; ywvia exaxepa<; xov evxoc xal aTtevavxiov ycoviCSv 
jieiCwv eaxiv. 

Tiaxco xpiytovov xo ABr, xal TtpoaexpepXria-dc) auxoO 
^ua TtXeupa f] Br era xo A- Xeyto, oxi f) exxo<; ywvta f) utio 
ArA ^lei^tov eaxlv exaxepac; xwv evxoc; xal drcevavxtov iSv 
utto TBA, BAr yovifiv. 

Tex^if|a , do f] Ar 8txa xaxa xo E, xal STU^eux'deiaa f] BE 
expepXf]a , dw in euiSeiai; era xo Z, xal xeta-dw xrj BE far) f) 
EZ, xal eTteCeux'S" f) Zr, xal Sir^M V] Ar eicl xo H. 

'Excel ouv iar] eaxlv f] (jiev AE xfj EI 1 , f) 8e BE xfj EZ, 8uo 
8f) at AE, EB 8uat xau; TE, EZ laai etalv exaxepa exaxepcr 
xal ycovia rj Otto AEB ywvta xfj Otto ZEr lot] eaxiv xaxa 
xopucpfjv yap- pdan; apa f) AB pdaei xfj ZT iar) eaxiv, xal xo 
ABE xpiytovov xu ZEr xpiycovcp eaxlv laov, xal at Xoircal 



For let the two straight-lines AB and CD cut one an- 
other at the point E. I say that angle AEC is equal to 
(angle) DEB, and (angle) CEB to (angle) AED. 
A 




B 



For since the straight-line AE stands on the straight- 
line CD, making the angles CEA and AED, the (sum 
of the) angles CEA and AED is thus equal to two right- 
angles [Prop. 1.13]. Again, since the straight-line DE 
stands on the straight-line AB, making the angles AED 
and DEB, the (sum of the) angles AED and DEB is 
thus equal to two right-angles [Prop. 1.13]. But (the sum 
of) CEA and AED was also shown (to be) equal to two 
right-angles. Thus, (the sum of) CEA and AED is equal 
to (the sum of) AED and DEB [C.N. 1]. Let AED have 
been subtracted from both. Thus, the remainder CEA is 
equal to the remainder BED [C.N. 3]. Similarly, it can 
be shown that CEB and DEA are also equal. 

Thus, if two straight-lines cut one another then they 
make the vertically opposite angles equal to one another. 
(Which is) the very thing it was required to show. 

Proposition 16 

For any triangle, when one of the sides is produced, 
the external angle is greater than each of the internal and 
opposite angles. 

Let ABC be a triangle, and let one of its sides BC 
have been produced to D. I say that the external angle 
ACD is greater than each of the internal and opposite 
angles, CBA and BAG. 

Let the (straight-line) AC have been cut in half at 
(point) E [Prop. 1.10]. And BE being joined, let it have 
been produced in a straight-line to (point) FJ And let 
EE be made equal to BE [Prop. 1.3], and let EC have 
been joined, and let AC have been drawn through to 
(point) G. 

Therefore, since AE is equal to EC, and BE to EF, 
the two (straight-lines) AE, EB are equal to the two 



20 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



ycovioa xalc; Xoutau; Yioviaic; Taai rialv sxaxspa exaxepa, ucp' 
at; ai Taai TtXeupal Ouoxeivouaiv Tar] apa saxlv f] utto BAE 
xrj U7i6 ErZ. jieiCwv 8£ eaxiv f] Otto ErA xrj<; Otto ETZ- 
^iriCcov apa f] Otto ATA xrj<; Otto BAE. 'O[io'ux>z Sr] xfjc; Br 
xex[ir]^£vr]<; Si/a Beix'Sffoexai xal f) Otto BrH, xouxsaxiv rj 
Otco ArA, urii^cov xal xfj? Otco ABr. 




Ilavxoc; apa xpiywvou ^iiac; xfiv TiXeupGv upooex- 
pXrj'vMarjc; f] exxog ywvia exaxepac; xfiv evxoc; xal dme- 
vavxiov ywviwv (jeiCuv eaxiv onep e8ei 8ric;ai. 



(straight-lines) Ci?, EF, respectively. Also, angle AEB 
is equal to angle FEC, for (they are) vertically opposite 
[Prop. 1.15]. Thus, the base AB is equal to the base FC, 
and the triangle ABE is equal to the triangle FEC, and 
the remaining angles subtended by the equal sides are 
equal to the corresponding remaining angles [Prop. 1.4]. 
Thus, BAE is equal to ECF. But ECD is greater than 
ECF. Thus, ACD is greater than BAE. Similarly, by 
having cut BC in half, it can be shown (that) BCG — that 
is to say, ACD — (is) also greater than ABC. 

A .F 




Thus, for any triangle, when one of the sides is pro- 
duced, the external angle is greater than each of the in- 
ternal and opposite angles. (Which is) the very thing it 
was required to show. 



t The implicit assumption that the point F lies in the interior of the angle ABC should be counted as an additional postulate. 



Ilavxoc; xpiywvou ai 860 ywviai 860 opiStov eXdaaovsc; 
rial Ttdvxfj ^exaXa^pavo^svai. 

A 




b r a 

"Eaxw xpiywvov xo ABT - Xeyco, oxi xou ABT xpiywvou 
ai 860 Y wv ' al §uo op^wv eXdxxovec; rial Ttdvxr] (jiexaXa^t- 
Pavojievai. 



Proposition 17 

For any triangle, (the sum of) two angles taken to- 
gether in any (possible way) is less than two right-angles. 

A 




B CD 

Let ABC be a triangle. I say that (the sum of) two 
angles of triangle ABC taken together in any (possible 
way) is less than two right-angles. 



21 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



'ExpepAr](Td<j yap f\ BT iia xo A. 

Kal tnei xpiyiovou tou ABr exxoc; eaxi y«via r\ bnb 
ATA, ^ei^iov eaxi xfjc; evxoc; xai drcevavxiov xfjc; bnb ABT. 
xoivf] npoaxeio'dw f) bub ATB- ai dpa Otto ArA, ArB x£>v 
Otto ABr, BEA [leiZovez eiaiv. dXX' ai Otto ArA, ArB 
860 op^dic; laai etaCv ai apa Otto ABr, BrA 860 op^wv 
eXdaaovec; eiaiv. o^ioicoc; 8f) 8eic;o^ev, oxi xai ai Otto BAr, 
ArB 860 op-dSv eXdaaovec eiai xai exi ai Otco TAB, ABT. 

navxoc; dpa xpiywvou ai 860 ywviai 860 op'dwv eXdac;- 
ovec; eiai Tidvxfj ^exaXa^pavo^evai- oTtep eBei 8eTc;ai. 



ir]'. 

navxoc; xpiycovou f] ^e[£wv TiXeupd xf]V ^ei^ova ywviav 
UTtoxdvei. 



A 




"Eaxw yap xpiycovov xo ABr [isiCova e)(ov xf)v Ar 
TtXeupdv xfjc AB- Xeyw, oxi xai ycovia f] Otto ABr (jlel^cov 
eaxi xfjc Otto BEA- 

Tkei yap [Jiei^cov eaxiv r] Ar xfjc AB, xeiai!)« xfj AB tar] 
f] AA, xai STieCe'JX'dw f] BA. 

Kai CTtei xpiyovou xoO BrA exxoc; eaxi ywvia f) Otto 
AAB, jieiCwv eaxi xfjc; evxoc; xai aTievavxiov xfjc; bnb ArB' 
iar] 8e f] Otto AAB xfj (mo ABA, ine\ xai TiXeupd f] AB xfj 
AA eaxiv laiy fiei^cov dpa xai f) bnb ABA xfjc; Otto ATB- 
ttoXXG apa f] Otto ABr ^.ei^wv eaxi xfjc; Otco ArB. 

navxoc; dpa xpiyovou f) [icii^wv TiXeupd xf)v ^.ei£ova 
ywviav UTioxeivei- ouep e8ei 8eTc;ai. 

Eiavxoc; xpiyiovou bnb xfjv ^ei^ova ywviav f\ ^teiC«v 
TiXeupd UTioxeivei. 

Tiaxco xpiycovov xo ABr ^leic^ova e)(ov xf]v Otco ABr 
yuviav xfjc; Otto BrA- Xeyw, 6x1 xai TiXeupd f] Ar TcXeupdc; 
xfjc; AB ^leiCwv eaxiv. 



For let BC have been produced to D. 

And since the angle ACD is external to triangle ABC, 
it is greater than the internal and opposite angle ABC 
[Prop. 1.16]. Let ACB have been added to both. Thus, 
the (sum of the angles) ACD and ACB is greater than 
the (sum of the angles) ABC and BCA. But, (the sum of) 
ACD and ACB is equal to two right-angles [Prop. 1.13]. 
Thus, (the sum of) ABC and BCA is less than two right- 
angles. Similarly, we can show that (the sum of) BAC 
and ACB is also less than two right-angles, and further 
(that the sum of) CAB and ABC (is less than two right- 
angles). 

Thus, for any triangle, (the sum of) two angles taken 
together in any (possible way) is less than two right- 
angles. (Which is) the very thing it was required to show. 

Proposition 18 

In any triangle, the greater side subtends the greater 
angle. 

A 




For let ABC be a triangle having side AC greater than 
AB. I say that angle ABC is also greater than BCA. 

For since AC is greater than AB, let AD be made 
equal to AB [Prop. 1.3], and let BD have been joined. 

And since angle ADB is external to triangle BCD, it 
is greater than the internal and opposite (angle) DCB 
[Prop. 1.16]. But ADB (is) equal to ABD, since side 
AB is also equal to side AD [Prop. 1.5]. Thus, ABD is 
also greater than ACB. Thus, ABC is much greater than 
ACB. 

Thus, in any triangle, the greater side subtends the 
greater angle. (Which is) the very thing it was required 
to show. 

Proposition 19 

In any triangle, the greater angle is subtended by the 
greater side. 

Let ABC be a triangle having the angle ABC greater 
than BCA. I say that side AC is also greater than side 
AB. 



22 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



Et yap \if\, fjxoi. iar) eaxlv f) Ar xfj AB fj eXdaawv tar] 
(lev ouv oux eaxiv f] Ar xfj AB- lor] yap dv rjv xod ytovia f] 
Otto ABr xfj (mo ArB- oux eaxi 8e- oux apa for) eaxlv f) Ar 
xfj AB. ou8e fjif|v eXdaawv eaxlv rj Ar xfj? AB- eXdaatov 
yap dv rjv xal ywvia f) utco ABr xfjc utio ArB- oux eaxi 
8e- oux apa eXdaatov eaxlv f\ AT xfj? AB. eSsix^^ 8s, oxi 
oG8e Tar] eaxiv. ^ie[£c>)v apa eaxlv f] Ar xfj? AB. 



A 



B 




r 

Ilavxo? apa xpiywvou Otto xrjv (jiei^ova ywviav f] (iclCwv 
icXeupa uiroxeivei- oiiep e8ei 8eT?"ai. 

X . 

Ilavxo? xpiycovou ai 8uo nXeupal xfj? XoiTtfj? ^ei£ove? 
eiai Tidvxr] ^exaXa^3av6^evai. 

A 




"Eaxw yap xpiywvov xo ABr- Xeyw, oxi xou ABr 
xpiyovou ai Suo uXeupal xfjc; Xomfj? ^lei^ove? eiai Tcavxri 
ti£xaXa[ipav6[i£vai, ai [ie\i BA, Ar xfj? Br, ai Be AB, Br 
xfj? Ar, ai Be Br, TA xfj? AB. 



For if not, AC is certainly either equal to, or less than, 
AB. In fact, AC is not equal to AB. For then angle ABC 
would also have been equal to ACB [Prop. 1.5]. But it 
is not. Thus, AC is not equal to AB. Neither, indeed, is 
AC less than AB. For then angle ABC would also have 
been less than ACB [Prop. 1.18]. But it is not. Thus, AC 
is not less than AB. But it was shown that (AC) is not 
equal (to AB) either. Thus, AC is greater than AB. 



A 



B 




C 

Thus, in any triangle, the greater angle is subtended 
by the greater side. (Which is) the very thing it was re- 
quired to show. 

Proposition 20 

In any triangle, (the sum of) two sides taken to- 
gether in any (possible way) is greater than the remaining 
(side). 

D 




B C 



For let ABC be a triangle. I say that in triangle ABC 
(the sum of) two sides taken together in any (possible 
way) is greater than the remaining (side). (So), (the sum 
of) BA and AC (is greater) than BC, (the sum of) AB 



23 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



Air))cdco yap fj BA em to A arj^eTov, xal xeiaftw xfj IA 
lot) f] AA, xal £Tie^£U)cd« f) Ar. 

'EtcsI ouv far) eaxlv f) A A xfj Ar, ia/) eaxl xal ywvia 
f) uno AAr xfj Otio ATA- jiei^wv apa f) utto BrA xfjc uno 
AAr- xal ETxel xpiywvov eaxi xo ArB (jieiCova e)(ov xrjv utio 
BrA ywviav xfjc; Otto BAr, Otto 8e xfjv jiei^ova ywviav f] 
^ei^wv TiXeupa Onoxeivei, f] AB apa xfjc; Br eaxi ^eic^wv. Tar) 
8e r) AA xfj Ar- jiei^ovec; apa al BA, Ar xfjc Br- ojioiwc; 
8f) Beii;o|jiev, oxi xal ai y.e\> AB, Br xfjc; TA [leiCovec eiaiv, 
ai 8e Br, FA xfjc; AB. 

Llavxoc; apa xpiywvou ai 860 TtXeupai xfjc; Xoiitfjc; 
[Lei^ovez eloi Tidvxr) uexaXa^pavo^evai- oitep e8ei Sel^ai. 



xa'. 

'Eav xpiywvou era (iiac; xwv TtXeupwv duo xwv Tiepdxwv 
860 eui£>eTai evxoc; auaxai!)waiv, ai auaxa-delaai xwv Xoitcwv 
xoO xpiywvou 860 TtXeupwv eXdxxovec; \ie\i eaovxai, ^tei^ova 
8e ywviav 7i:epiec;ouaiv. 




Tpiywvou yap xoO ABr era ^iiac; xwv nXeupwv xfjc; Br 
dn:6 xwv Tiepdxwv xwv B, T 860 eui9eTai evxoc; auveaxdxwaav 
ai BA, Ar- Xeyw, oxi ai BA, Ar xwv Xoittwv xou xpiywvou 
860 TtXeupwv xwv BA, Ar eXdaaovec; [lev eiaiv, [iei^ova 8s 
ywviav Tispisxouoi xfjv utio BAr xfjc; bub BAT. 

Aif|)cdw yap f] BA era xo E. xal CTtel Ttavxoc xpiywvou 
ai 860 TtXeupai xfjc; XoiTtfjc; (jiei^ovec; eiaiv, xoO ABE apa 
xpiywvou ai 860 TtXeupai ai AB, AE xfjc; BE ^tei^ovec 
eiaiv xoivf) Ttpoaxeia'dw f) EE ai apa BA, Ar xwv BE, 
Er jiei^ovec; eiaiv. TtdXiv, CTtel xou TEA xpiywvou ai 860 
TtXeupai ai TE, EA xfjc; EA [iciCovec; eiaiv, xoivf] Ttpoaxeia'dw 
f] AB- ai EE, EB apa xwv EA, AB jieiCovec; eiaiv. dXXa 
xwv BE, Er [leiCovec; e8ei)edr)aav ai BA, Ar- TtoXXw apa ai 
BA, Ar xwv BA, Ar ^d^ovec; eiaiv. 

ndXiv, eitei Ttavxoc; xpiywvou f) exxoc; ywvia xfjc; evxoc; 
xal aTtevavxiov jieii^wv eaxiv, xou EAE apa xpiywvou f) 
exxoc; ywvia f) bub BAT ^lei^wv eaxi xfjc; utio TEA. 81a 
xauxa xoivuv xal xou ABE xpiywvou f] exxoc; ywvia f) bub 



and BC than AC, and (the sum of) BC and CM than 
AB. 

For let have been drawn through to point D, and 
let AD be made equal to CA [Prop. 1.3], and let DC 
have been joined. 

Therefore, since DA is equal to AC, the angle ADC 
is also equal to ACD [Prop. 1.5]. Thus, BCD is greater 
than ADC. And since DCB is a triangle having the angle 
BCD greater than BDC, and the greater angle subtends 
the greater side [Prop. 1.19], DB is thus greater than 
BC. But DA is equal to AC. Thus, (the sum of) BA and 
AC is greater than BC. Similarly, we can show that (the 
sum of) AB and BC is also greater than CA, and (the 
sum of) and C.4 than AB. 

Thus, in any triangle, (the sum of) two sides taken to- 
gether in any (possible way) is greater than the remaining 
(side). (Which is) the very thing it was required to show. 

Proposition 21 

If two internal straight-lines are constructed on one 
of the sides of a triangle, from its ends, the constructed 
(straight-lines) will be less than the two remaining sides 
of the triangle, but will encompass a greater angle. 




For let the two internal straight-lines BD and DC 
have been constructed on one of the sides BC of the tri- 
angle ABC, from its ends B and C (respectively) . I say 
that BD and DC are less than the (sum of the) two re- 
maining sides of the triangle BA and AC, but encompass 
an angle BDC greater than BAG. 

For let BD have been drawn through to E. And since 
in any triangle (the sum of any) two sides is greater than 
the remaining (side) [Prop. 1.20], in triangle ABE the 
(sum of the) two sides AB and AE is thus greater than 
BE. Let EC have been added to both. Thus, (the sum 
of) BA and AC is greater than (the sum of) BE and EC. 
Again, since in triangle CED the (sum of the) two sides 
CE and ED is greater than CD, let DB have been added 
to both. Thus, (the sum of) CE and EB is greater than 
(the sum of) CD and DB. But, (the sum of) BA and 
AC was shown (to be) greater than (the sum of) BE and 
EC. Thus, (the sum of) BA and AC is much greater than 



24 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



TEB (jiei^wv iav. tfj? Otto BAI\ dXXd xfjc; uno TEB (jiei^wv 
sSeix'&T) T) utto BAr- ttoXXG apa f] uno BAr ^lei^wv eaxi 
xfjc Oko BAr. 

'Edv apa xpiyGvou era [iidc; xGv TtXeupGv dno xSv 
Ttepdxcov Buo eu-delai evxo<; auaxa'dwaiv, ai auaxa'deTaai x£>v 
XoitcGv xou xpiyGvou Buo TtXeupGv eXdxxovec; jiev eiaiv, 
^iei£ova Be ywviav nepiexouaiv oitep eBei SeT^ai. 



(the sum of) BD and DC. 

Again, since in any triangle the external angle is 
greater than the internal and opposite (angles) [Prop. 
1.16], in triangle CDE the external angle BDC is thus 
greater than CED. Accordingly, for the same (reason), 
the external angle CEB of the triangle ABE is also 
greater than BAC. But, BDC was shown (to be) greater 
than CEB. Thus, BDC is much greater than BAC. 

Thus, if two internal straight-lines are constructed on 
one of the sides of a triangle, from its ends, the con- 
structed (straight-lines) are less than the two remain- 
ing sides of the triangle, but encompass a greater angle. 
(Which is) the very thing it was required to show. 



x(3'. 

'Ex xpiGv eu'deiwv, off eiaiv Xoou. xpioi xau; So'deiaau; 
[eOifteiau;] , xpiyovov auox^aaCTdai - Set Be xd<; 860 xfjc Xomfjc 
(jidi^ovac; eTvai Ttdvxy] (iexaXa^pavo^eva<; [Bid xo xai navxoc 
xpiyovou xac; Buo rcXeupac xfjc Xomfjt; [iei^ovac eTvai Tidvxr] 
(i£xaXa^i[3avou£vac;] . 



Proposition 22 

To construct a triangle from three straight-lines which 
are equal to three given [straight-lines]. It is necessary 
for (the sum of) two (of the straight-lines) taken together 
in any (possible way) to be greater than the remaining 
(one), [on account of the (fact that) in any triangle (the 
sum of) two sides taken together in any (possible way) is 
greater than the remaining (one) [Prop. 1.20] ]. 

A 

B 

C 













z 


H 




e 




'Eoxwaav ai Bo-deTaai xpeTc; euiMai ai A, B, T, Sv ai 
Buo xfjc Xomfjc \LZiC,ovec, eoxwoav Kavxr) (lexaXa^pavo^ievai, 
ai ]izv A, B xfj<; T, ai Be A, T xrj<; B, xal exi ai B, T xfjc A- 
Bel Br) ex xGv lawv xdic; A, B, T xpiywvov auaxriaacrdai. 

'Exxeicrdco xic; eMela f] AE KCTiepaajievr) [lev xaxd xo 
A aTteipoc; Be xaxd xo E, xai xeicrdco xfj [ism A lot] f] AZ, 
xfj Be B tar) f] ZH, xfj Be T Tar) rj HO- xai xevxptp jiev xG 
Z, Biaaxrj^iaxi Be iS ZA xuxXo<; yeypdcpTdw 6 AKA - TtdXiv 
xevxpcp |iev xG H, Biaaxr^iaxi Be xG H6 xuxXoc; yeypdcp-dco 
6 KA0, xai ejie^eu)cd«aav ai KZ, KH- Xeyco, oxi ex xpiGv 
eMeiGv xGv lacov xoii<; A, B, T xpiywvov ouveaxaxai xo 
KZH. 

Tkei yap xo Z arjueTov xevxpov eaxi xou AKA xuxXou, 
iar] eaxiv f) ZA xfj ZK- dXXd f] ZA xfj A eaxiv lar\. xai f) 



D 













F 


G 




H 




Let A, B, and C be the three given straight-lines, of 
which let (the sum of) two taken together in any (possible 
way) be greater than the remaining (one). (Thus), (the 
sum of) A and B (is greater) than C, (the sum of) A and 
C than £?, and also (the sum of) B and C than A. So 
it is required to construct a triangle from (straight-lines) 
equal to A, B, and C. 

Let some straight-line DE be set out, terminated at D, 
and infinite in the direction of E. And let DF made equal 
to A, and FG equal to 5, and GH equal to C [Prop. 1.3]. 
And let the circle DKL have been drawn with center F 
and radius FD. Again, let the circle KLH have been 
drawn with center G and radius GH. And let KF and 
KG have been joined. I say that the triangle KFG has 



25 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



KZ apa xfj A eaxiv Xar\. TidXiv, ETtd to H g/^eTov xevxpov 
eaxl xoO AK0 xuxXou, lor] eaxlv #] H9 xfj HK- dXXa f] H9 
xfj T saxiv laiy xal f\ KH apa xfj T eaxiv lor), eaxl Be xal f] 
ZH xfj B tar) - ai xpdc; apa eCWteTai ai KZ, ZH, HK xpial xaic 
A, B, r i'aai daiv. 

'Ex xpiGv apa eu-daGv xGv KZ, ZH, HK, ou. daiv 
laai xpial xaic; BoiJeiaaic eCWteiau; xau; A, B, T, xpiycovov 
auveaxaxai xo KZH oicep e§ei itoifjaai. 



been constructed from three straight-lines equal to A, B, 
and C. 

For since point F is the center of the circle DKL, FD 
is equal to FK. But, FD is equal to A. Thus, KF is also 
equal to A. Again, since point G is the center of the circle 
LKH, GH is equal to GK. But, GH is equal to C. Thus, 
KG is also equal to C. And FG is also equal to B. Thus, 
the three straight-lines KF, FG, and GA are equal to A, 
B, and C (respectively). 

Thus, the triangle KFG has been constructed from 
the three straight-lines KF, FG, and GK, which are 
equal to the three given straight-lines A, B, and C (re- 
spectively) . (Which is) the very thing it was required to 
do. 



xy . 

npoc; xfj Bo-deia/] eO'dda xal xG Ttpoc; auxfj or\[Lsicd 

xfj 5oi9dar] yovia eO'duypa^.u.to I'arjv ywviav EU'duypa^ov 
auaxrpacrdai. 




Proposition 23 

To construct a rectilinear angle equal to a given recti- 
linear angle at a (given) point on a given straight-line. 




z 



H B 

"Eaxco f\ \ie\> Bcdeiaa eO'dda fj AB, xo Be Ttpoc; auxfj 
arjueTov xo A, f] 8e 8o$daa ywvia eO'duypa^oc; f] Oico ArE- 
8eT 6f] TCpoc; xfj Scdeiar) eui&eia xfj AB xal xG npoc, auxfj 
ar)udcp xG A xfj 8oi9e[ar] ywvia eG^uypd^iJiip xfj uito ArE 
'larjv ycoviav einJuypaijijj.ov auaxfpacrdai. 

EiX^cpdo ecp' exaxepac; xGv TA, TE xuxovxa arj^ida xa 
A, E, xal eice^eux'dco f) AE- xal ex xpiGv eui5eiGv, ai eiaiv 
law. xpial xaic; IA, AE, TE, xpiycovov auveaxdxco xo AZH, 
Gaxe larjv elvai x/jv [lev TA xfj AZ, xf]v 8s TE xfj AH, xal 
exi xf)v AE xfj ZH. 

'End ouv Buo ai AT, TE 86o xaic; ZA, AH I'aai dalv 
exaxepa exaxepa, xal pdau; f] AE pdaei xfj ZH lot], yovia 
dpa f] utco ArE ywvia xfj utco ZAH eaxiv iarj. 

npoc; dpa xfj Bo'deiar] eu-deia xfj AB xal xG npoc; auxfj 
arjueiw xG A xfj Bo'deiar] ywvia eu'duypd^w xfj utco ArE 
Xar] yovia euiJuypa^oc; auveaxaxai f) utco ZAH- oTtep eBei 
Ttoirjaai. 



F 




G B 



Let AB be the given straight-line, A the (given) point 
on it, and DCE the given rectilinear angle. So it is re- 
quired to construct a rectilinear angle equal to the given 
rectilinear angle DCE at the (given) point A on the given 
straight-line AB. 

Let the points D and E have been taken at random 
on each of the (straight-lines) CD and CE (respectively), 
and let DE have been joined. And let the triangle AFC 
have been constructed from three straight-lines which are 
equal to CD, DE, and CE, such that CD is equal to AF, 
CE to AG, and further DE to FG [Prop. 1.22]. 

Therefore, since the two (straight-lines) DC, CE are 
equal to the two (straight-lines) FA, AG, respectively, 
and the base DE is equal to the base FG, the angle DCE 
is thus equal to the angle FAG [Prop. 1.8]. 

Thus, the rectilinear angle FAG, equal to the given 
rectilinear angle DCE, has been constructed at the 
(given) point A on the given straight-line AB. (Which 




2G 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



x5'. 

'Edv Buo xpiytova xdc; Buo nXeupdc; [xdic;] 860 TtXeupaTc; 
I'aac; eyr\ exaxepav exaxepa, xrjv Be ytovlav xfjc; ytovlac; 
^e[£ova ex?) xr ] v utto T " v i° wv eu-deiwv Tiepie)(o^evr]v, xal 
xf]v pdaiv Tfj? pdaetoc; ^lel^ova ec;ei. 

A A 



B \ \/7 



"Eaxco Buo xpiywva xd ABr, AEZ xdc; Buo TtXeupdc; 
xdc; AB, Ar xaic; Buo TtXeupaTc; xaic; AE, AZ I'aac; exovxa 
exaxepav exaxepa, xrjv |iev AB xfj AE xf]v Be Ar xfj AZ, f) 
Be Ttpoc; xa> A ycovia xfjc; Tipoc; xai A ycoviac; ^xei^cov eaxw 
Xeyco, 6x1 xal pdaig f] Br pdaewc; xfjc; EZ ^ei^wv eaxiv. 

Tkel y«P ^ei^wv f) Otto BAr ywvia xfjc; utio EAZ 
ycoviac;, auveaxdxw Ttpoc; xrj AE eu-deia xal x65 Tipoc; auxfj 
a/jjieicp xw A xfj Otto BAr ycovia Xat] i\ utto EAH, xal xeiai9« 
OTtoxepa xwv Ar, AZ lay] f) AH, xal CTie^eu)(i9«oav ai EH, 
ZH. 

Tkel ouv Xat] eaxlv f] [Lev AB xfj AE, f] Be Ar xrj AH, 
Buo Br) ai BA, Ar Buai xaic; EA, AH i'aai eialv exaxepa 
exaxepa- xal yiovia f] utio BAr ywvia ^fi utio EAH Xai]- 
pdaic; apa f) Br pdaei xrj EH eaxiv Tar). TidXiv, CTiel Xat] 
eaxlv f| AZ xfj AH, Tar) eaxl xal f] Otto AHZ ycovia xfj utio 
AZH' [ici^wv apa f) utio AZH xfjc utio EHZ' TioXXai apa 
[ici^iov eaxlv f) utio EZH xfjc; Otto EHZ. xal ercel xpiycovov 
eaxi xo EZH jiet^ova e/ov xf]v utio EZH ywviav xfjc; utio 
EHZ, utio Be xrjv jiet^ova ywviav f) [leii^v TtXeupd UTraxeivei, 
^eiCwv apa xal TtXeupd i\ EH xfjc; EZ. Xor] Be f) EH xfj BE 
^ei^tov apa xal f) Br xfjc; EZ. 

'Eav apa 860 xpiywva xdc; Buo TtXeupdc; Bual TtXeupaTc; 
laac; syr\ exaxepav exaxepa, xf)v Be ywvlav xfjc; ywviac; 
^eii^ova exT) xfjv utco xGv tawv eu^eiOv Ttepiexo^ievr]v, xal 
xf)v pdaiv xfjc; pdaewc; ^tei^ova e^er oTtep eBei Bel^ai. 



is) the very thing it was required to do. 

Proposition 24 

If two triangles have two sides equal to two sides, re- 
spectively, but (one) has the angle encompassed by the 
equal straight-lines greater than the (corresponding) an- 
gle (in the other), then (the former triangle) will also 
have a base greater than the base (of the latter) . 



A D 




Let ABC and DEF be two triangles having the two 
sides AB and AC equal to the two sides DE and DF, 
respectively. (That is), AB (equal) to DE, and AC to 
DF. Let them also have the angle at A greater than the 
angle at D. I say that the base BC is also greater than 
the base EF. 

For since angle BAC is greater than angle EDF, 
let (angle) EDG, equal to angle BAC, have been 
constructed at the point D on the straight-line DE 
[Prop. 1.23]. And let DC be made equal to either of 
AC or DF [Prop. 1.3], and let EG and FG have been 
joined. 

Therefore, since AB is equal to DE and AC to DC, 
the two (straight-lines) BA, AC are equal to the two 
(straight-lines) ED, DG, respectively. Also the angle 
BAC is equal to the angle EDG. Thus, the base BC 
is equal to the base EG [Prop. 1.4]. Again, since DF 
is equal to DG, angle DGF is also equal to angle DFG 
[Prop. 1.5]. Thus, DFG (is) greater than EGF . Thus, 
EFG is much greater than EGF. And since triangle 
EFG has angle EFG greater than EGF, and the greater 
angle is subtended by the greater side [Prop. 1.19], side 
EG (is) thus also greater than EF. But EG (is) equal to 
BC. Thus, BC (is) also greater than EF. 

Thus, if two triangles have two sides equal to two 
sides, respectively, but (one) has the angle encompassed 
by the equal straight-lines greater than the (correspond- 
ing) angle (in the other), then (the former triangle) will 
also have a base greater than the base (of the latter). 




27 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



xz. 

'Eav 860 xpiywva xdc 860 TtXeupdc. 8uol TtXeupdlc laac 
zyj\ exaxepav exaxepa, xf)v 8e pdaiv xfjc. pdaewc. ^.ei^ova 
e/r), xal x/)v yioviav xfjc ycoviac. ^ei^ova ei;ei xf]v utto xfiv 
i'ouv eui)eifiv Ttepiexojievrjv. 



A 




'Eaxco 860 xpiywva xd ABr, AEZ xdc. 860 TtXeupdc. 
xdc. AB, Ar xalc 860 TtXeupaTc xalc AE, AZ I'aac. exovxa 
exaxepav exaxepa, xrjv jiev AB xfj AE, xf|v 6e Ar xfj AZ- 
pdaic. 8e f] BT pdaeoc xfj? EZ ue(C«v eaxw Xey«, oxi xal 
ycovia f] utio BAr ycoviac xfjc utio EAZ jiei^wv eaxiv. 

EE yap [ir|, fjxoi Tar) eaxiv auxfj rj eXdaatov Tar) ^tev ouv 
oux eaxiv f] utio BAT xfj Otto EAZ- lay) yap av rjv xal pdaic 
f] Br pdaei xfj EZ- oux eaxi 8e. oux apa Tar) eaxi ycovia f] 
utio BAr xrj utio EAZ- ouSe |if]v eXdaawv eaxiv f] utio BAr 
xfjc utio EAZ- eXdaatov yap av rjv xal pdaic. f] Br pdaeGJc. 
xfjc EZ- oux ecru 6e- oux apa eXdaawv eaxiv f) utio BAr 
ycovia xfjc. utio EAZ. eSeL/iS/] 8e, 6x1 ouSe Tar)- ^eii^tov apa 
eaxiv f| utio BAr xfjc utio EAZ. 

'Eav apa 860 xpiycova xdc. 860 TiXeupdc. 8ual TrXeupdic. 
1'aac. exfj exaxepav exaxepa, xf]v 8e Paaiv xfjc. pdaetoc 
[ici^ova sxTli Tr ] v ywviav xfjc, yioviac. jiei^ova e^ei xfjv 
uno xcov 'lacov eui5eifiv Tiepie/o(jievr]V- oiiep e8ei SeT^ai. 



Xf'. 

'Eav Suo xpiyova xdc. 860 ycoviac. Sual ycoviaic. 1'aac. 'iyjl 
exaxepav exaxepa xal puav TtXeupdv ^iia TiXeupa iar]v fjxoi xf)V 
Tipoc. xalc laaic ycoviaic fj xfjv UTtoxeivouaav utio ^uav iSv 
Tacov ytoviaiv, xal xdc XoiTidc. TiXeupdc. xalc; XoiTidic. TiXeupdic. 
1'aac. e5ei [exaxepav exaxepa] xal xfjv XoiTifjv yioviav xfj XoiTifj 
ycovia. 

TSaxco 860 xpiycova xd ABr, AEZ xdc. 8uo ycoviac. xdc 



(Which is) the very thing it was required to show. 

Proposition 25 

If two triangles have two sides equal to two sides, 
respectively, but (one) has a base greater than the base 
(of the other), then (the former triangle) will also have 
the angle encompassed by the equal straight-lines greater 
than the (corresponding) angle (in the latter) . 



A 




E F 

Let ABC and DEF be two triangles having the two 
sides AB and AC equal to the two sides DE and DF, 
respectively (That is), AB (equal) to DE, and AC to DF. 
And let the base BC be greater than the base EF. I say 
that angle BAG is also greater than EDF. 

For if not, (BAC) is certainly either equal to, or less 
than, (EDF). In fact, BAC is not equal to EDF. For 
then the base BC would also have been equal to the base 
EF [Prop. 1.4]. But it is not. Thus, angle BAC is not 
equal to EDF . Neither, indeed, is BAC less than EDF . 
For then the base BC would also have been less than the 
base EF [Prop. 1.24]. But it is not. Thus, angle BAC is 
not less than EDF. But it was shown that (BAC is) not 
equal (to EDF) either. Thus, BAC is greater than EDF . 

Thus, if two triangles have two sides equal to two 
sides, respectively, but (one) has a base greater than the 
base (of the other), then (the former triangle) will also 
have the angle encompassed by the equal straight-lines 
greater than the (corresponding) angle (in the latter). 
(Which is) the very thing it was required to show. 

Proposition 26 

If two triangles have two angles equal to two angles, 
respectively, and one side equal to one side — in fact, ei- 
ther that by the equal angles, or that subtending one of 
the equal angles — then (the triangles) will also have the 
remaining sides equal to the [corresponding] remaining 
sides, and the remaining angle (equal) to the remaining 
angle. 



28 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



0x6 ABr, BrA 5ual xau; 0tt:6 AEZ, EZA laac; eyrovxa 
exaxepav exaxepa, xf)v ^tev 0x6 ABr xfj 0x6 AEZ, xf)v 
8s Otio BEA xfj 0x6 EZA- exexco Se xal ^iav TtXeupdv [iia 
xXeupa larjv, Ttpoxepov xfjv upoc; xalc; laai? ytavtaic; xf)v 
Br xfj EZ- Xeyco, oxi xal xd<; Xoixac; xXeupac; xalc; Xoixdic; 
xXeupau; laa? ec;ei exaxepav exaxepa, xf)v ^tev AB xfj AE 
xf]v 8s Ar xfj AZ, xal xf)V Xoixf)v ycovlav xfj XoiTtfj ycovla, 
xfjv 0x6 BAr xfj 0x6 EAZ. 




El yap dviaoc; eaxiv f) AB xfj AE, aOxcov ^telCcov 
eaxiv. eaxco ^el^cov f) AB, xal xeladco xfj AE lot) f) BH, xal 
exci^eu^-dco f] HT. 

'EtceI ouvl'or) eoxlv f) jiev BH xfj AE, f) 8s Br xfj EZ, 660 
8f) al BH, Br 6ual xalc; AE, EZ I'oai eiolv exaxepa exaxepa- 
xal ycovla f] 0x6 HBT ycovla xfj 0x6 AEZ I'or) eoxlv pdaic; 
apa f) Hr pdaei xfj AZ I'or) eaxiv, xal xo HBr xplycovov xco 
AEZ xpiycovcp Taov eoxlv, xal al Xoixal ycovlai xau; Xoixdic; 
ycovlaic; Taai eoovxai, Ocp' ac; al Taai TiXeupal OTtoxelvouoiv 
Tar) apa f) Otto HrB ycovla xfj 0x6 AZE. dXXa f) 0x6 AZE 
xfj Otto BEA UTtoxeixai Tar)' xal f) 0x6 BrH apa xfj Otto BEA 
Tar) eoxlv, f) eXdaocov xfj jiel^ovi- oxep dBuvaxov. oOx dpa 
dviaoc; eaxiv f) AB xfj AE. I'or) dpa. eaxi 8e xal f) Br xfj EZ 
Tar)' 860 Sf) al AB, Br 8uol xalc; AE, EZ Taai eiolv exaxepa 
exaxepa' xal ycovla f) 0x6 ABr ycovla xfj 0x6 AEZ eaxiv 
Tar)' pdoig apa f) Ar pdaei xfj AZ I'or) eaxiv, xal Xoixf) ycovla 
f) 0x6 BAr xfj Xoixfj ycovla xfj O716 EAZ I'or) eoxlv. 

AXXd Sf) TidXiv eaxcoaav al Otto xdc; Taac; ycovlac; xXeupal 
Oxoxelvouaai I'oai, cbc; f) AB xfj AE' Xeyco xdXiv, oxi xal al 
Xomal xXeupal xalc; Xoixdic; TiXeupdic; Taai eoovxai, f) (lev Ar 
xfj AZ, f) 8e Br xfj EZ xal exi f) Xomf) ycovla i\ 0x6 BAr 
xfj XoiTtfj ycovla xfj Otto EAZ Tar) eaxiv. 

Ei yap dviooc; eaxiv f) Br xfj EZ, [ila aOxcov jiel^cov eaxiv. 
eaxco ^elc^cov, ei 8uvax6v, f) Br, xal xeladco xfj EZ Tar) f) B9, 
xal CTie^e0)(i9co fj A6. xal ctcci Tor) eoxlv f) ^iev B9 xfj EZ 
f) 8e AB xfj AE, 660 Sf) al AB, B6 8ual xalc; AE, EZ Taai 
eiolv exaxepa exapepa- xal ycovlac; Taac; xepiexouaiv pdoic; 
apa f] AO pdaei xfj AZ Tar) eoxlv, xal xo AB9 xplycovov xco 
AEZ xpiycovcp Taov eoxlv, xal al Xoixal ycovlai xalc; Xoixdic; 
ycovlaic; I'oai eoovxai, Ocp' ac; al Taac; TtXeupal O-xoxelvouaiv 
lor) apa eaxiv f) 0x6 B0A ycovla xfj 0x6 EZA. dXXa f) Otto 



Let ABC and DEF be two triangles having the two 
angles ABC and BCA equal to the two (angles) DEF 
and EFD, respectively. (That is) ABC (equal) to DEF , 
and BCA to EFD. And let them also have one side equal 
to one side. First of all, the (side) by the equal angles. 
(That is) BC (equal) to EF. I say that they will have 
the remaining sides equal to the corresponding remain- 
ing sides. (That is) AB (equal) to DE, and AC to DF. 
And (they will have) the remaining angle (equal) to the 
remaining angle. (That is) BAC (equal) to EDF. 



D 




For if AB is unequal to DE then one of them is 
greater. Let AB be greater, and let BG be made equal 
to DE [Prop. 1.3], and let GC have been joined. 

Therefore, since BG is equal to DE, and BC to EF, 
the two (straight-lines) GB, BC^ are equal to the two 
(straight-lines) DE, EF, respectively. And angle GBC is 
equal to angle DEF. Thus, the base GC is equal to the 
base DF, and triangle GBC is equal to triangle DEF, 
and the remaining angles subtended by the equal sides 
will be equal to the (corresponding) remaining angles 
[Prop. 1.4]. Thus, GCB (is equal) to DFE. But, DFE 
was assumed (to be) equal to BCA. Thus, BCG is also 
equal to BCA, the lesser to the greater. The very thing 
(is) impossible. Thus, AB is not unequal to DE. Thus, 
(it is) equal. And BC is also equal to EF. So the two 
(straight-lines) AB, BC are equal to the two (straight- 
lines) DE, EF, respectively. And angle ABC is equal to 
angle DEF. Thus, the base AC is equal to the base DF, 
and the remaining angle BAC is equal to the remaining 
angle EDF [Prop. 1.4]. 

But, again, let the sides subtending the equal angles 
be equal: for instance, (let) AB (be equal) to DE. Again, 
I say that the remaining sides will be equal to the remain- 
ing sides. (That is) AC (equal) to DF, and BC to EF. 
Furthermore, the remaining angle BAC is equal to the 
remaining angle EDF. 

For if BC is unequal to EF then one of them is 
greater. If possible, let BC be greater. And let BH be 
made equal to EF [Prop. 1.3], and let AH have been 
joined. And since BH is equal to EF, and AB to DE, 
the two (straight-lines) AB, BH are equal to the two 



29 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



EZA xfj utio BEA eaxiv Tar) - xpiywvou 8f] xou AGE f] exxoc 
ywvia f] Otto BOA iar) eaxi xfj evxoc; xai dracvavxiov xfj Otto 
BEA- oitep dSuvaxov. oux dpa dviaoc; eaxiv f\ BE xfj EZ- lor\ 
dpa. eaxl Se xai f] AB xfj AE iar). 8uo Sf] ai AB, BE 8uo 
xdic; AE, EZ laai eialv exaxepa exaxepa- xai yoviac; laac; 
Tiepiexouar Pdaic; dpa f) Ar pdaei xfj AZ iar) eaxiv, xai xo 
ABE xpiytovov iu AEZ xpiywvtp laov xai Xomf] yovia f] 
utio BAE xfj Xoircf] yovia xfj utio EAZ far). 

'Edv dpa 860 xpiycova xac Suo ywvtac; 8uoi yoviaic; laac 
exaxepav exaxepa xai ^tiav TiXeupav [j.ia TiXeupd i'arjv 
fjxoi xfjv Ttpoc; xdic; Taaic; ycoviaic;, *) xr)V UTioxeivouaav Otto 
(jiiav xwv Tacov ywviwv, xai xac; Xomdc; nXeupag xdic; XoiTtdic 
TtXeupdic; laac; e$ei xai xf|V Xomf]V ytoviav xfj Xomfj ycoviqr 
oTiep e8ei 8eTc;ai. 



t The Greek text has "BG, BC", which is obviously a mistake. 

'Edv etc; 860 eu-fJeiac; eu'de'ia epiraTtxouaa xdc; evaXXac; 
ywvia; laa; dXXiqXai; Ttoifj, TiapdXXrjXoi eaovxai dXXr]Xai; ai 
eMelai. 




H 



El? yap 860 eMeia; xac AB, EA eui5ela e^iraTixouaa f] 
EZ xdc; evaXXd? ycovia; xdc; utio AEZ, EZA i'aa; dXXrjXai; 
Ttoieixoy Xeyw, oxi TtapdXXrjXo; eaxiv f) AB xfj EA. 

EE yap [if], ex(3aXX6^£vai ai AB, EA au^Tieaouvxai rjxoi 
E7xi xd B, A |iepr) fj era xd A, E. expepXiqaOwaav xai ouji- 
rajtxexcoaav era xd B, A (iepr) xaxd xo H. xpiywvou 8f) xou 
HEZ f) exxo; ywvia f] utio AEZ iar) eaxi xfj evxoc; xai arce- 
vavxiov xfj utio EZH- oTtep eaxiv dSuvaxov oux dpa ai AB, 
AE expaXXojievai au^TteaoOvxai era xd B, A [iepr]. ojioiio; 



(straight-lines) DE, EF, respectively. And the angles 
they encompass (are also equal). Thus, the base AH is 
equal to the base DF, and the triangle ABH is equal to 
the triangle DEF, and the remaining angles subtended 
by the equal sides will be equal to the (corresponding) 
remaining angles [Prop. 1.4]. Thus, angle BHA is equal 
to EFD. But, EFD is equal to BCA. So, in triangle 
AHC, the external angle BHA is equal to the internal 
and opposite angle BCA. The very thing (is) impossi- 
ble [Prop. 1.16]. Thus, BC is not unequal to EF. Thus, 
(it is) equal. And AB is also equal to DE. So the two 
(straight-lines) AB, BC are equal to the two (straight- 
lines) DE, EF, respectively. And they encompass equal 
angles. Thus, the base AC is equal to the base DF, and 
triangle ABC (is) equal to triangle DEF, and the re- 
maining angle BAG (is) equal to the remaining angle 
EDF [Prop. 1.4]. 

Thus, if two triangles have two angles equal to two 
angles, respectively, and one side equal to one side — in 
fact, either that by the equal angles, or that subtending 
one of the equal angles — then (the triangles) will also 
have the remaining sides equal to the (corresponding) re- 
maining sides, and the remaining angle (equal) to the re- 
maining angle. (Which is) the very thing it was required 
to show. 



Proposition 27 

If a straight-line falling across two straight-lines 
makes the alternate angles equal to one another then 
the (two) straight-lines will be parallel to one another. 




G 



For let the straight-line EF, falling across the two 
straight-lines AB and CD, make the alternate angles 
AEF and EFD equal to one another. I say that AB and 
CD are parallel. 

For if not, being produced, AB and CD will certainly 
meet together: either in the direction of B and D, or (in 
the direction) of A and C [Def. 1.23]. Let them have 
been produced, and let them meet together in the di- 
rection of B and D at (point) G. So, for the triangle 



30 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



8f) 8ei)fdf|aexai, oxi ou8e em xd A, E at 8s era (ir]8exepa xd 
^tepr] aujiTUTixouaai TcapdXXrjXot etmv TiapdXXr)Xoc; apa eaxiv 
#1 AB xrj rA. 

'Edv apa etc; 8uo eu'deiac; eu'dela ejiTUTCxouaa xdc; evaXXdi; 
ywvtac; faac; dXXfjXaic; Tioifj, TiapdXXr]Xoi eaovxai at euiSelai- 
oTiep e8ei 8eTc;ai. 



XT]'. 

'Edv etc; 8uo ei/deiac; eO'deia ejirarcxouaa xf|v exxoc; 
ywvtav xrj evxog xal dnevavxtov xat era xd auxa \iipt) far)v 
Tioifj fj xdc; evxoc; xal era xd auxa [ispr] Sualv opdal? faac, 
TiapdXXrjXoi Eaovxai dXXfjXaic; at eO'delai. 




Etc; yap 60o ei/deiac; xa<; AB, IA eui5eTa ejiraTcxouaa f] 
EZ xf|v exxoc; ycoviav xf]v utco EHB xfj evxoc; xal dnevavxbv 
yovta xfj utio HOA \'ar)v tcoicixgj fj xdc; evxoc; xat stil xd 
auxa [iepi] xdc; utio BH0, H0A §uatv op'dau; faac; - Xeyto, 
oxi TiapdXXr)X6<; eaxiv f] AB xrj TA. 

'EticI yap far] eaxiv f) utio EHB xrj utco HOA, dXXd f] utco 
EHB xrj Otto AH0 eaxiv far), xal f) utio AH0 apa xrj utio 
H6A eaxiv far) - xat etaiv evaXXd^' TcapdXXr]Xoc; apa eaxtv f) 
AB xrj TA. 

ndXiv, eitet at utco BH0, H8A 8uo op'dalc; taai etaiv, 
etat 8e xat at utco AH8, BH0 8uatv opiate; t'aai, at apa 
utco AH0, BH0 xalc; utco BH0, H0A taai etaiv xoivf] 
dcpr]pf)adw f] utio BH6 - Xomf] apa f] utco AH0 XoiTcfj xrj 
utco HOA eaxiv far) - xat eiaiv evaXXd^ - TcapdXXrjXoc apa 
eaxiv f) AB xrj TA. 

'Edv apa etc; Suo eu'deiac; eu'deta ejiTciTcxouaa xf)v exxoc; 
ycoviav xrj evxoc; xat aTcevavxtov xat era xd auxa [ispt] farjv 



GEF, the external angle AEF is equal to the interior 
and opposite (angle) EFG. The very thing is impossible 
[Prop. 1.16]. Thus, being produced, AB and CD will not 
meet together in the direction of B and D. Similarly, it 
can be shown that neither (will they meet together) in 
(the direction of) A and C. But (straight-lines) meeting 
in neither direction are parallel [Def. 1.23]. Thus, AB 
and CD are parallel. 

Thus, if a straight-line falling across two straight-lines 
makes the alternate angles equal to one another then 
the (two) straight-lines will be parallel (to one another) . 
(Which is) the very thing it was required to show. 

Proposition 28 

If a straight-line falling across two straight-lines 
makes the external angle equal to the internal and oppo- 
site angle on the same side, or (makes) the (sum of the) 
internal (angles) on the same side equal to two right- 
angles, then the (two) straight-lines will be parallel to 
one another. 




For let EF, falling across the two straight-lines AB 
and CD, make the external angle EGB equal to the in- 
ternal and opposite angle GHD, or the (sum of the) in- 
ternal (angles) on the same side, BGH and GHD, equal 
to two right-angles. I say that AB is parallel to CD. 

For since (in the first case) EGB is equal to GHD, but 
EGB is equal to AGH [Prop. 1.15], AGH is thus also 
equal to GHD. And they are alternate (angles). Thus, 
AB is parallel to CD [Prop. 1.27] . 

Again, since (in the second case, the sum of) BGH 
and GHD is equal to two right-angles, and (the sum 
of) AGH and BGH is also equal to two right-angles 
[Prop. 1.13], (the sum of) AGH and BGH is thus equal 
to (the sum of) BGH and GHD. Let BGH have been 
subtracted from both. Thus, the remainder AGH is equal 
to the remainder GHD. And they are alternate (angles). 
Thus, AB is parallel to CD [Prop. 1.27]. 



31 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



Ttoifj fj xd? evxo? xal era xd auxd [iepr) 8ualv op'ddi? Taa?, 
TiapdXXr)Xoi eaovxai otl eu'dslai- oracp s8ei Bsi^ai. 



X0'. 

H d? xag TiapaXXf|Xou? euifteia? sMeia ejiraTixouaa xd? 
te evaXXdc; ywvia? taa? dXXrjXai? Traiei xal xfjv sxxo? xfj 
svxo? xal dtTievavTiov larjv xal xd? evxo? xal era xa auxd 
y.spf] Sualv op'ddi? Taa?. 




El? yap TiapaXXf|Xou? eui9da? xd? AB, EA su-deia 
s^tutitetco f] EZ- Xsyco, oxi xd? evaXXd?" ytovia? xd? Otto 
AH9, H9A Taa? Tioiei xal xf|v exxo? yioviav xf)v Otto EHB 
XT] evxo? xal dmsvavxbv xfj Otto H9A Tar]v xal xa? svxo? 
xal stxI xd auxd [iepf] xd? utio BH6, H9A Sualv opi&al? 
Taa?. 

EE yap aviao? saxiv f) utio AH9 xfj utio H9A, \i{<x auxcov 
iiei^wv saxiv. Eaxco [lei^cov f) utio AH9- xoivf] Tipooxeiai9« 
f] utio BH9- ai apa utio AH9, BH9 xwv Otto BH9, H9A 
[idZoMSQ eiaiv. dXXd at utio AH9, BH9 Sualv op'ddi? Taai 
slow, [xal] ai apa utio BH9, H9A Suo 6p$Gv sXdaaove? 
riaiv. ai §£ dm' eXaaaovwv f] Suo opiSwv expaXXo^ievai 
si? aTisipov aupuibixouaiv ai apa AB, EA ixpaXXo^ievai 
si? draipov au^Tisaouvxai- ou au^iraTixouai Se Bid xo Tia- 
paXXf|Xou? auxd? OiioxeTadai- oux apa dviao? eaxiv f) Otto 
AH9 xrj utio H9A- i'ar) apa. dXXd f] utio AH9 xfj Otto EHB 
eaxiv Tary xal f] utio EHB apa xfj Otto H9A eaxiv tar) - xoivf] 
Tipoaxeia'do f) Otto BH9" ai apa utio EHB, BH9 xal? utio 
BH9, H9A Taai eiaiv. dXXd ai utio EHB, BH9 Suo opflal? 
Taai eiaiv xal ai Otto BH9, H9A apa Suo op'ddi? Taai eiaiv. 

C H apa ei? xd? TtapaXXf|Xou? eOiJeia? eu'dela e^traTixouaa 
xd? xe evaXXd?" ywvia? Taa? dXXf]Xai? TioieT xal xfjv exxo? 
xfj evxo? xal aTievavxiov Tarjv xal xd? evxo? xal era xd auxd 



Thus, if a straight-line falling across two straight-lines 
makes the external angle equal to the internal and oppo- 
site angle on the same side, or (makes) the (sum of the) 
internal (angles) on the same side equal to two right- 
angles, then the (two) straight-lines will be parallel (to 
one another). (Which is) the very thing it was required 
to show. 

Proposition 29 

A straight-line falling across parallel straight-lines 
makes the alternate angles equal to one another, the ex- 
ternal (angle) equal to the internal and opposite (angle), 
and the (sum of the) internal (angles) on the same side 
equal to two right-angles. 




For let the straight-line EF fall across the parallel 
straight-lines AB and CD. I say that it makes the alter- 
nate angles, AGH and GHD, equal, the external angle 
EGB equal to the internal and opposite (angle) GHD, 
and the (sum of the) internal (angles) on the same side, 
BGH and GHD, equal to two right-angles. 

For if AGH is unequal to GHD then one of them is 
greater. Let AGH be greater. Let BGH have been added 
to both. Thus, (the sum of) AGH and BGH is greater 
than (the sum of) BGH and GHD. But, (the sum of) 
AGH and BGH is equal to two right-angles [Prop 1.13]. 
Thus, (the sum of) BGH and GHD is [also] less than 
two right-angles. But (straight-lines) being produced to 
infinity from (internal angles whose sum is) less than two 
right-angles meet together [Post. 5]. Thus, AB and CD, 
being produced to infinity, will meet together. But they do 
not meet, on account of them (initially) being assumed 
parallel (to one another) [Def. 1.23]. Thus, AGH is not 
unequal to GHD. Thus, (it is) equal. But, AGH is equal 
to EGB [Prop. 1.15]. And EGB is thus also equal to 
GHD. Let BGH be added to both. Thus, (the sum of) 
EGB and BGH is equal to (the sum of) BGH and GHD. 
But, (the sum of) EGB and BGH is equal to two right- 



32 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



[iepr] 8uoiv opdal? !aac oTcep e5ei 8eT?ai. angles [Prop. 1.13]. Thus, (the sum of) BGff and GHD 

is also equal to two right-angles. 

Thus, a straight-line falling across parallel straight- 
lines makes the alternate angles equal to one another, the 
external (angle) equal to the internal and opposite (an- 
gle), and the (sum of the) internal (angles) on the same 
side equal to two right-angles. (Which is) the very thing 
it was required to show. 



X'. 

Ai xfj auxfj eu'dsla TcapdXXr]Xoi xod dXXr|Xai<; eial TcapdXXr]- 

Xoi. 




Tiaxco sxaxspa x£>v AB, FA xfj EZ TcapdXXrjXoc;- Xey", 
on xod r) AB xrj TA eraxi TiapdXXrjXoc;. 

'E^7iitix£xcl) yap slz auxag eu'deTa f] HK. 

Kod enel tic, TcapaXXf|Xou<; eu-Mac; xd<; AB, EZ su-dela 
e^KEnxcdxev f\ HK, tar] dpa f] utio AHK xfj utco H9Z. 
raxXiv, eitel eu; TcapaXXf|Xou<; euiDeia^ xd<; EZ, EA eu'dela 
e^KETixcdxev f\ HK, 'far) eaxlv f) utio H0Z xrj utco HKA. 
e8B)edr) 8e xal f] utio AHK xfj utco H9Z lor], xod f\ utco AHK 
dpa xfj utio HKA eaxiv I'ar)- xoa eiaiv evaXXd^. TiapdXXr]Xo<; 
dpa eaxlv f] AB xfj FA. 

[Ai dpa xfj auxfj eu-Ma TcapdXXr]Xoi xal dXXr|Xai<; rial 
jcapdXXr]Xoi-] OTcep e8ei 8ric;ai. 



Proposition 30 

(Straight-lines) parallel to the same straight-line are 
also parallel to one another. 




Let each of the (straight-lines) AB and CD be parallel 
to EF. I say that AB is also parallel to CD. 

For let the straight-line GK fall across {AB, CD, and 
EF). 

And since the straight-line GK has fallen across the 
parallel straight-lines AB and EF, (angle) AGK (is) thus 
equal to GHF [Prop. 1.29]. Again, since the straight-line 
GK has fallen across the parallel straight-lines EF and 
CD, (angle) GHF is equal to GKD [Prop. 1.29]. But 
AGK was also shown (to be) equal to GHF. Thus, AGK 
is also equal to GKD. And they are alternate (angles). 
Thus, AB is parallel to CD [Prop. 1.27]. 

[Thus, (straight-lines) parallel to the same straight- 
line are also parallel to one another.] (Which is) the very 
thing it was required to show. 



Xa'. 

Aid xou 8oi9£vxo<; ar\{ieiou xfj Sorrier) eu-dsia uapdXXrjXov 
su-driav ypa^fjv dyayriv. 

'Eaxw xo y.ev Scdev arjueTov xo A, i] Se Bcdriaa eu-dria 
f) Br- Sri Sf] Sid xou A ar^dou xfj BT eu'deia TiapdXXrjXov 
su-driav ypaji^fjv dyayriv. 

ElXfjcp-dw etc! xfj? BT xu)(6v arjuriov xo A, xal £TieC£U)fdw 
f] AA- xal auveaxdxco Tipog xfj AA eui9ria xal xw Tip6<; auxfj 
arjfieicp xw A xfj utio AAr y«v(a lor\ f\ utco AAE- xal 



Proposition 31 

To draw a straight-line parallel to a given straight-line, 
through a given point. 

Let A be the given point, and BC the given straight- 
line. So it is required to draw a straight-line parallel to 
the straight-line BC, through the point A. 

Let the point D have been taken a random on BC, and 
let AD have been joined. And let (angle) DAE, equal to 
angle ADC, have been constructed on the straight-line 



33 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



sxpspXiqaiSw etc' eO'delac; xfj EA eu-Qsla f) AZ. 




Kal etcei etc; 8uo eu'deiac; xac; Br, EZ eu'dsTa e^nuTixouaa 
f] A A xac; evaXXdc; ytoviac; xac; utio EAA, AAr I'aac; 
dXXf]Xaic; TteTio(r]xev, TtapdXXr]Xoc; apa eaxlv f) EAZ xfj Br. 

Aid xoO 8o , devxoc; apa ar^eiou xoO A xfj Bo'deiar] cu-dela 
xfj Br TiapdXXrjXoc; eui9eTa ypa^turj ^ XTai ^ EAZ- oTtep eBei 
noifjaai. 



DA at the point A on it [Prop. 1.23]. And let the straight- 
line AF have been produced in a straight-line with EA. 




And since the straight-line AD, (in) falling across the 
two straight-lines BC and EF, has made the alternate 
angles EAD and ADC equal to one another, EAF is thus 
parallel to BC [Prop. 1.27]. 

Thus, the straight-line EAF has been drawn parallel 
to the given straight-line BC, through the given point A. 
(Which is) the very thing it was required to do. 



A(3'. 

Ilavxoc; xpiyovou [iiac; xQv TiXeupwv TipoaexpXrj'Marjc; 
f] exxoc; yiovia 8uol xau; evxoc; xal aTievavxiov lot) eaxlv, xod 
al evxoc; xoO xpiywvou xpsTc ycoviai 5ualv op'dalc; iaai eiaiv. 



A E 




B T A 

*Ectto xpiytovov xo ABr, xal TcpoaexpepXrja-dc) auxoO 
[lia TiXeupd f] Br era. xo A- Xeyw, oxi f] exxoc; ywvia f] utio 
ArA tar) eaxl 8ual xalc; evxoc; xal aTievavxiov xalc; utio TAB, 
ABr, xal ai evxoc; xoO xpiytovou xpeTc; ytovlai al utio ABr, 
BrA, TAB Bualv op^dic; iaai eiaiv. 

"Hy^co yap Bid xoO T arj^elou xfj AB eO'deia TiapdXXrjXoc; 
r I K. 

Kal etcsi TiapdXXrjXoc; eaxiv fj AB xfj TE, xal eic; auxdc; 
ejiTieKxcoxev f\ AT, al evaXXdi; ycovlai al utio BAr, ATE iaai 
dXXrjXaic; eiaiv. udXiv, enel TiapdXXrjXoc; eaxiv f) AB xfj TE, 
xal eic; auxdc; e^TieTixwxev eu-dela f) BA, f] exxoc; ytovla f) 
utio ErA iar) eaxl xfj evxoc; xal aTievavxiov xfj utio ABr. 
eSei/i}/) 5e xal f) utio ArE xfj utio BAr Tor)- oXrj dpa f\ utio 
ArA ycovia Iar) eaxl 8ual xdic; evxoc; xal aTievavxiov xalc; utco 

BAr, ABr. 



Proposition 32 

In any triangle, (if) one of the sides (is) produced 
(then) the external angle is equal to the (sum of the) two 
internal and opposite (angles), and the (sum of the) three 
internal angles of the triangle is equal to two right-angles. 
A E 




BCD 



Let ABC be a triangle, and let one of its sides BC 
have been produced to D. I say that the external angle 
ACD is equal to the (sum of the) two internal and oppo- 
site angles CAB and ABC, and the (sum of the) three 
internal angles of the triangle — ABC, BCA, and CAB — 
is equal to two right-angles. 

For let CE have been drawn through point C parallel 
to the straight-line AB [Prop. 1.31]. 

And since AB is parallel to CE, and AC has fallen 
across them, the alternate angles BAC and ACE are 
equal to one another [Prop. 1.29]. Again, since AB is 
parallel to CE, and the straight-line BD has fallen across 
them, the external angle ECD is equal to the internal 
and opposite (angle) ABC [Prop. 1.29]. But ACE was 
also shown (to be) equal to BAC. Thus, the whole an- 



34 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



Koivf) TtpoaxsitrdtL) f\ Otto ATB- at apa utco ArA, ArB 
xpial xau; Otio ABr, BrA, TAB l'aai eiaiv. dXX' at bub ATA, 
ArB Sualv opflau; iaai eiaiv xal at utco ArB, TBA, TAB 
apa Sualv 6pi9aT<; laai eknv. 

Ilavxoc; apa xpiytovou ^iiac; xfiv xXeupfiv xpoaex- 
pXr^Bar^ f\ exxoc; ytovia Bual xaic; evxoc; xal dxevavxiov 
lar) eaxiv, xal at evxoc; xou xpiytovou xpeu; ywviai 8ualv 
opiate; laai eiaiv oxep eSei 8eTc;ai. 



gle ACD is equal to the (sum of the) two internal and 
opposite (angles) BAG and ABC. 

Let ACB have been added to both. Thus, (the sum 
of) ACD and ACB is equal to the (sum of the) three 
(angles) ABC, BCA, and CAB. But, (the sum of) ACL> 
and ACB is equal to two right-angles [Prop. 1.13]. Thus, 
(the sum of) ACB, CBA, and CAB is also equal to two 
right-angles. 

Thus, in any triangle, (if) one of the sides (is) pro- 
duced (then) the external angle is equal to the (sum of 
the) two internal and opposite (angles), and the (sum of 
the) three internal angles of the triangle is equal to two 
right-angles. (Which is) the very thing it was required to 
show. 



Ay. 

At xd<; laac; xe xal 7tapaXXf|Xouc; era. xd auxd ^epr] era- 
i^euyvuouaai euiJeTai xal auxal laai xe xal xapdXX/]Xoi eiaiv. 

B A 




A r 

"Eaxcoaav laai xe xal raxpdXX/jXoi at AB, TA, xal exi- 
i^euyvuxoaav aOxdc; era xd auxd uepr) eu-delai at AT, BA- 
Xeyo, oxi xal at Ar, BA laai xe xal xapdXXr)Xo[ eiaiv. 

'Eue^eux'dw f] Br. xal eitet xapdXXr]X6c; eaxiv f] AB xfj 
TA, xal etc auxd<; eu-XCTixoxev f] Br, at evaXXdi; yoviai at 
uxo ABr, BrA laai dXXf|Xai<; eiaiv. xal exel Xor\ eaxlv f) AB 
xfj TA xoivf] 8e f] Br, 8uo Bf] at AB, Br 8uo xaic; Br, TA laai 
eiaiv xal ywvia f] bub AST ywvia xfj 0x6 BrA lary pdau; 
apa f] AT pdaei xfj BA eaxiv iar), xal xo ABr xpiyovov xcp 
BrA xpiywvw laov eaxiv, xal at Xoixal ywviai xaic; XoixaTc; 
ywviaic; laai eaovxai exaxepa exaxepa, Ocp' ac; at laai xXeupal 
O^oxetvouaiv iar) apa f] bub ATB yovia xfj 0x6 TBA. xal 
exel etc; 8uo eOiSeiac; xdc; AT, BA eO-dela ejixixxouaa f) Br 
xdc; evaXXdc; ywviac; laac; dXXfjXaic; xexo(/]xev, xapdXXrjXoc; 
apa eaxlv f] AT xfj BA. eSeix^r] 8e auxfj xal iar). 

At apa xdc; laac; xe xal 7iapaXXf]Xouc; era xd auxd \±£pr\ 
exii^euyvuouaai euiJeTai xal auxal laai xe xal xapdXX/]Xo( 
eiaiv Sxep e8ei 8eTc;ai. 



Proposition 33 

Straight-lines joining equal and parallel (straight- 
lines) on the same sides are themselves also equal and 
parallel. 

B A 




D C 

Let AB and CD be equal and parallel (straight-lines), 
and let the straight-lines AC and BD join them on the 
same sides. I say that AC and BD are also equal and 
parallel. 

Let BC have been joined. And since AB is paral- 
lel to CD, and BC has fallen across them, the alter- 
nate angles ABC and BCD are equal to one another 
[Prop. 1.29]. And since AB is equal to CD, and BC 
is common, the two (straight-lines) AB, BC are equal 
to the two (straight-lines) DC, CBJ And the angle ABC 
is equal to the angle BCD. Thus, the base AC is equal 
to the base BD, and triangle ABC is equal to triangle 
DCB^, and the remaining angles will be equal to the 
corresponding remaining angles subtended by the equal 
sides [Prop. 1.4]. Thus, angle ACB is equal to CBD. 
Also, since the straight-line BC, (in) falling across the 
two straight-lines AC and BD, has made the alternate 
angles {ACB and CBD) equal to one another, AC is thus 
parallel to BD [Prop. 1.27]. And (AC) was also shown 
(to be) equal to (BD). 

Thus, straight-lines joining equal and parallel (straight- 



35 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



t The Greek text has "BC, CD", which is obviously a mistake. 
* The Greek text has "DCB", which is obviously a mistake. 

X8'. 

Tc5v 7iapaXXr]Xoypd[j.[jiiov ^topiwv al aTievavxiov TiXeupal 
xe xal ywvlai faai dXXr|Xaic; eiaiv, xal f\ Bidjiexpoc; auxa Si/a 
xe^vei. 

A B 




r a 

"Egxcl> 7tapaXX/]X6ypa[i^ov ^wpiov xo ArAB, Bid^iexpoc; 
8e auxou f) Br- Xeyw, oxi xou ArAB TtapaXXrjXoypdjijjiou al 
dnevavxlov TiXeupal xe xal yovlai laai dXXr|Xai<; eiaiv, xal f] 
Br Bidjiexpoc; auxo 8()(a xe^tvei. 

Tkel yap napdXXr]X6<; eaxiv 7] AB xfj TA, xal sic, auxdg 
e^TCETixcoxEV euiDeTa rj Br, al evaXXai; ycovlai a ' 1 U7I ° ABr, 
BrA Taai aXX^Xaic; elalv. TtdXiv enel TiapdXXrjXog eaxiv f] Ar 
xfj BA, xal ei? auxdc; e^mcTixcoxev r] Br, al evaXXai; ycovlai 
al Otco ArB, TBA Taai dXXr|Xai<; elalv. 8uo 8r) xplycovd eaxi 
xd ABr, BrA xdc; 8uo ywviac; xd<; Otto ABr, BrA 8ual 
xdlc; bub BrA, TBA I'aag e)(ovxa exaxepav exaxepa xal ^tlav 
TtXeupdv [iia TtXeupa Tarjv x/]v Ttpoc; xdlc; Taai? ywviaic; xoivrjv 
auxov x/]v Br- xal xdc; Xomdc; apa TtXeupdc; xdlc; XoiTtdlc; 
Taac; ec;ei exaxepav exaxepa xal xrjv Xoircrjv ycovlav xfj XoiTtfj 
ywvia- I'ar) apa f) [lev AB TtXeupd xfj TA, f] 8e Ar xfj BA, 
xal exi Xat] eaxlv r] bub BAT yovla xfj Otio TAB. xal ensl 
Tar] eaxlv f] ^iev O716 ABr ycovia xfj bub BrA, rj 8e bub TBA 
xfj Oko ArB, oXr) apa f) Otco ABA oXr] xfj 0n:6 ArA eaxiv 
Xar]. eBelx'dr) 8e xal f] bub BAT xfj Otto TAB Tar]. 

Twv apa 7tapaXXr]Xoypd^a>v x w P^ wv a 'i aTievavxiov 
TiXeupal xe xal ywviai I'aai dXXrjXaic; elalv. 

Aeyw 8t], 6x1 xal f] Sid^texpoc; auxa Sixa xefivei. enel yap 
Tar] eaxlv f) AB xfj TA, xoivrj 8e f] Br, 860 Bf] al AB, Br 
8ual xdlc; TA, Br Taai elalv exaxepa exaxepa- xal ywvla f) 
Otto ABr ytovla xfj Otto BrA la/], xal pdaic; apa f] AT xfj 
AB Tar], xal xo ABr [apa] xplycovov xw BrA xpiyovo Taov 
eaxlv. 

C H apa Br 8id(iexpo<; 8l)(a xejivei xo ABrA uapaX- 
Xf]kbypa.[i[Lov onep e8ei 8ele;ai. 



lines) on the same sides are themselves also equal and 
parallel. (Which is) the very thing it was required to 
show. 



Proposition 34 

In parallelogrammic figures the opposite sides and angles 
are equal to one another, and a diagonal cuts them in half. 



A B 




C D 



Let ACDB be a parallelogrammic figure, and BC its 
diagonal. I say that for parallelogram ACDB, the oppo- 
site sides and angles are equal to one another, and the 
diagonal BC cuts it in half. 

For since AB is parallel to CD, and the straight-line 
BC has fallen across them, the alternate angles ABC and 
BCD are equal to one another [Prop. 1.29]. Again, since 
AC is parallel to BD, and BC has fallen across them, 
the alternate angles ACB and CBD are equal to one 
another [Prop. 1.29]. So ABC and BCD are two tri- 
angles having the two angles ABC and BCA equal to 
the two (angles) BCD and CBD, respectively, and one 
side equal to one side — the (one) by the equal angles and 
common to them, (namely) BC. Thus, they will also 
have the remaining sides equal to the corresponding re- 
maining (sides), and the remaining angle (equal) to the 
remaining angle [Prop. 1.26]. Thus, side AB is equal to 
CD, and AC to BD. Furthermore, angle BAG is equal 
to CDB. And since angle ABC is equal to BCD, and 
CBD to ACB, the whole (angle) ABD is thus equal to 
the whole (angle) ACD. And BAC was also shown (to 
be) equal to CDB. 

Thus, in parallelogrammic figures the opposite sides 
and angles are equal to one another. 

And, I also say that a diagonal cuts them in half. For 
since AB is equal to CD, and BC (is) common, the two 
(straight-lines) AB, BC are equal to the two (straight- 
lines) DC, CB\ respectively. And angle ABC is equal to 
angle BCD. Thus, the base AC (is) also equal to DB, 



3G 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



and triangle ABC is equal to triangle BCD [Prop. 1.4]. 

Thus, the diagonal BC cuts the parallelogram ACDB^ 
in half. (Which is) the very thing it was required to show. 

t The Greek text has "CD, BC", which is obviously a mistake. 
* The Greek text has "ABCD", which is obviously a mistake. 



Xe'. 

Ta TtapaXXrjXoYpa^a xd era xfjc; aGxfjc; pdoewc; ovxa xal 
ev xalc; auxalc; TtapaXXfjXoic; Taa dXXf|Xoi<; eoxiv. 

A ' A E Z 




b r 

"Eaxw 7iapaXX/]X6Ypa[i^a xd ABTA, EBTZ era xfjc; 
auxfjc; pdoewc; xfjc; BT xal ev xau; auxau; TtapaXXrjXoic; xau; 
AZ, BT- Xsyw, oxi ioov laxl xo ABTA xG EBTZ TtapaXXr)- 
XoYpd^i^w. 

'Etcei Ydp napaXXiqXoYpa^ov eoxi xo ABTA, Xar\ eoxlv 
f] AA xfj BT. 8id xd auxd 8f) xal f) EZ xfj BT eoxiv for] - 
Goxe xal #] AA xfj EZ eoxiv for) - xal xoivf] f] AE- oXrj dpa 
fj AE oXr) xfj AZ eoxiv for), eoxi 5e xal f] AB xfj AT for) - 
8uo 8f) ai EA, AB Buo xau; ZA, AT I'aai eialv exaxepa 
exaxepa- xal yc^Via f) utto ZAT Y^via xfj bub EAB eoxiv 
I'or) T) exxoc; xfj evxoc;- pdou; dpa f] EB pdoei xfj ZT for) eoxiv, 
xal xo EAB xpiYiovov xG AZT xpiYGvip I'oov eoxai- xoivov 
dcprjprjadco xo AHE- Xoittov dpa xo ABHA xpaTte^iov Xomcp 
xG EHTZ xpanei^co eoxlv I'oov xoivov npoaxdadco xo HBT 
xpiYWvov 6Xov dpa xo ABTA 7iapaXXr)XoYpa|jiu.ov oXcp xG 
EBTZ 7rapaXXr)XoYpd^« I'oov eoxiv. 

Ta dpa TtapaXXr]X6Ypajj.[jia xd era xfjc; auxfjc; pdoewc; ovxa 
xal ev xalc; auxalc; 7tapaXXf|Xou; foa dXXfjXoic; eoxiv ouep e8ei 
SeT^ai. 



Proposition 35 

Parallelograms which are on the same base and be- 
tween the same parallels are equal^ to one another. 
A D E F 




B C 



Let ABCD and EBCF be parallelograms on the same 
base BC, and between the same parallels AF and BC. I 
say that ABCD is equal to parallelogram EBCF. 

For since ABCD is a parallelogram, AD is equal to 
BC [Prop. 1.34]. So, for the same (reasons), EF is also 
equal to BC. So AD is also equal to EF. And DE is 
common. Thus, the whole (straight-line) AE is equal to 
the whole (straight-line) DF. And AB is also equal to 
DC. So the two (straight-lines) EA, AB are equal to 
the two (straight-lines) FD, DC, respectively. And angle 
FDC is equal to angle EAB, the external to the inter- 
nal [Prop. 1.29]. Thus, the base EB is equal to the base 
FC, and triangle EAB will be equal to triangle DFC 
[Prop. 1.4]. Let DGE have been taken away from both. 
Thus, the remaining trapezium ABCD is equal to the re- 
maining trapezium EGCF. Let triangle GBC have been 
added to both. Thus, the whole parallelogram ABCD is 
equal to the whole parallelogram EBCF. 

Thus, parallelograms which are on the same base and 
between the same parallels are equal to one another. 
(Which is) the very thing it was required to show. 



t Here, for the first time, "equal" means "equal in area", rather than "congruent". 



X<r'. 

Td napaXXrjXoYpa^a xd era focov pdoewv ovxa xal ev 
xalc; auxalc; napaXXfjXoic; foa dXXfjXoic; eoxiv. 

"Eot(x> 7iapaXX/]X6Ypa[i^a xd ABTA, EZH6 era focov 
pdoewv ovxa xGv BT, ZH xal ev xalc; auxalc; 7tapaXXf|Xou; 
xalc; A0, BH- Xcy«, oxi I'oov eaxi xo ABTA TtapaX- 



Proposition 36 

Parallelograms which are on equal bases and between 
the same parallels are equal to one another. 

Let ABCD and EFGH be parallelograms which are 
on the equal bases BC and FG, and (are) between the 
same parallels AH and BG. I say that the parallelogram 



37 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



XrjXoYpaji^ov xcp EZH9. 

A A E 




BT Z H 

'ETieCeu)fd«oav yap od BE, TO. xal ctccI iarj eaxlv f] 
Br xfj ZH, dXXd f] ZH xfj E0 eaxiv Tar), xal f) Br dpa xfj 
E9 eaxiv Xar\. eial 8s xal napdXXrjXoi. xal era^euyvuouaiv 
auxdc ai EB, OI 1, ai 8s xdc iaac xe xal TtapaXXfjXouc era 
xd auxd (iepr) era^euyvuouaai I'aai xe xal uapdXX/jXoi eiai 
[xal al EB, QT dpa Taai xe rim xal TiapdXXr)Xoi] . uapaX- 
Xr]X6ypa[i|jiov dpa eaxl xo EBr6. xai eaxiv I'aov ifi ABTA- 
pdaiv xe yap auxcp xfjv aOxfjv exei xf|v Br, xal ev xdic auxalc 
TtapaXXf]Xoic eaxlv auxc5 xau; Br, A0. 8la xd auxd 8rj xal xo 
EZH6 ifi auxcp xw EBr0 eaxiv I'aov waxe xal xo ABrA 
TtapaXXr]X6ypa[i(jiov ifi EZH6 eaxiv i'aov. 

Td dpa TtapaXXr]X6ypa[j.[jia xd era lawv pdaewv ovxa xal 
ev xalc auxalc; 7tapaXXf|Xoic laa dXXf|Xoic eaxiv orcep e8ei 
8eu;ai. 



ABCD is equal to EFGH. 



AD EH 




B C F G 



For let BE and CH have been joined. And since BC is 
equal to FG, but FG is equal to EH [Prop. 1.34], BC is 
thus equal to EH. And they are also parallel, and EB and 
i?C join them. But (straight-lines) joining equal and par- 
allel (straight-lines) on the same sides are (themselves) 
equal and parallel [Prop. 1.33] [thus, EB and HC are 
also equal and parallel]. Thus, EBCH is a parallelogram 
[Prop. 1.34], and is equal to ABCD. For it has the same 
base, BC, as {ABCD), and is between the same paral- 
lels, BC and AH, as {ABCD) [Prop. 1.35]. So, for the 
same (reasons), EFGH is also equal to the same (par- 
allelogram) EBCH [Prop. 1.34]. So that the parallelo- 
gram ABCD is also equal to EFGH. 

Thus, parallelograms which are on equal bases and 
between the same parallels are equal to one another. 
(Which is) the very thing it was required to show. 



AC- 

Td xpiywva xd era xfjc auxfjc pdaewc ovxa xal ev xalc; 
auxalc napaXXfjXoic laa dXXfjXoic eaxiv. 

A A 




b r 

TCaxco xpiyova xd ABT, ABr era xfjc; auxfjc; pdaeoc xfjc 
Br xal ev xalc; auxalc TtapaXXf|Xoic xdic AA, BT- Xeyo, oxi 
laov eaxl xo ABT xpiywvov x£> ABT xpiyova). 

'ExpepXfja'dM f] AA ecp' exdxepa xd [lipf] era xd E, Z, xal 
8id \ism xou B xfj FA 7iapdXXr]Xoc fjx'dw f] BE, 8la 8e xou T xfj 
BA jiapdXXr]Xoc fj)edtL> f\ TZ. 7tapaXX/]X6ypa[i^.ov dpa eaxlv 
exdxepov xwv EBFA, ABTZ- xai riaiv iaa - era xe yap xfjc 
auxfjc pdaewc eloi xfjc Br xal ev xdic auxalc TtapaXXfjXoic 
xdic Br, EZ' xai eaxi xou ^.ev EBTA 7tapaXX/]Xoypd^ou 
fj^iiau xo ABT xpiywvov f] yap AB 8id[iexpoc auxo 8[)(a 
xe^vei- xou 8e ABTZ 7tapaXX/]Xoypd^ou fj[iiau xo ABr 
xpiycovov f) yap AT Sidjiexpoc auxo 8[)(a xejivei. [xd 8e 



Proposition 37 

Triangles which are on the same base and between 
the same parallels are equal to one another. 




B C 



Let ABC and DBC be triangles on the same base BC, 
and between the same parallels AD and BC. I say that 
triangle ABC is equal to triangle DBC. 

Let AD have been produced in both directions to E 
and F, and let the (straight-line) BE have been drawn 
through B parallel to CA [Prop. 1.31], and let the 
(straight-line) CF have been drawn through C parallel 
to BD [Prop. 1.31]. Thus, EBCA and DBCF are both 
parallelograms, and are equal. For they are on the same 
base BC, and between the same parallels BC and EF 
[Prop. 1.35]. And the triangle ABC is half of the paral- 
lelogram EBCA. For the diagonal AB cuts the latter in 



38 



ETOLXEIftN a'. 



ELEMENTS BOOK 1 



x£>v Tacov i]\iiay] Taa dXXr|Xoic; eaxiv] . Taov apa saxi to ABr 
xplywvov iw ABr xpiywvw. 

Td apa xplywva xd era xfj<; auxfj<; pdae«<; ovxa xal ev xau; 
auxau; 7iapaXXr|Xoi<; Taa dXXr]Xoi<; eaxiv oTtep eSei 8eTc;ai. 



t This is an additional common notion. 

XT)'. 

Td xpiyiova xd Era Tacov pdaecov ovxa xal ev xau; auxau; 
TtapaXXrjXou; Taa dXXrjXou; eaxiv. 

H A A 




BT E Z 

TEaxw xpiycova xd ABr, AEZ era Tacov pdaetov xwv Br, 
EZ xai ev xalg auxau; TtapaXXiqXou; xau; BZ, AA- Xeyto, oxi 
Taov eaxl xo ABr xpiyiovov xfi AEZ xpiyiovcp. 

'ExpepXrja'dco yap f) AA ecp' exdxepa xd jiepr] era xd H, 
0, xai Bid |iev xou B xf) TA TtapdXXr]Xoc; f])fdw f] BH, 8la 8s 
xou Z xfj AE 7iapdXX/]Xo<; f))fdw f] Z0. 7tapaXX/]X6ypa^ov 
apa eaxlv exdxepov xwv HBrA, AEZ6- xal Taov xo HBrA 
iS AEZ9- era xe yap Taov pdaewv eiai xov Br, EZ xal 
ev xau; auxau; TtapaXXr|Xou; xau; BZ, HO' xai eaxi xou [iev 
HBrA TtapaXXrjXoypd^iou r^uiau xo ABr xplywvov. f\ yap 
AB Bid^texpoc; auxo 8()(a xe^lvei' xou Be AEZ0 TtapaXXr]- 
Xoypd[i^tou 7]^iau xo ZEA xpiywvov rj yap AZ S(a^expo<; 
auxo 8()(a xe^tvei [xd Be xtov Tawv i]\±ioY) Taa aXXVjXou; eaxiv]. 
Taov apa eaxl xo ABr xplywvov xo AEZ xpiywvw. 

Td apa xpiywva xd era Tawv pdaetov ovxa xal ev xau; 
auxau; TtapaXXrjXou; Taa dXXf|Xoic; eaxiv oracp eSei 8eT<;ai. 



Td Taa xpiywva xd era xfj? auxrj<; pdaeioc; ovxa xal era 
xa auxd uepr) xal ev xau; auxau; 7iapaXXr|Xou; eaxiv. 

'Eaxco Taa xpiywva xd ABr, ABr era xfj? auxfjg pdae«<; 
ovxa xal era xd auxd [iepi] xfj? Br- Xeyw, oxi xal ev xau; 



half [Prop. 1.34]. And the triangle DBC (is) half of the 
parallelogram DBCF. For the diagonal DC cuts the lat- 
ter in half [Prop. 1.34]. [And the halves of equal things 
are equal to one another.] t Thus, triangle ABC is equal 
to triangle DBC. 

Thus, triangles which are on the same base and 
between the same parallels are equal to one another. 
(Which is) the very thing it was required to show. 



Proposition 38 

Triangles which are on equal bases and between the 
same parallels are equal to one another. 

G A D H 




B C E F 



Let ABC and DEF be triangles on the equal bases 
BC and EF, and between the same parallels BF and 
AD. I say that triangle ABC is equal to triangle DEF. 

For let AD have been produced in both directions 
to G and H, and let the (straight-line) BG have been 
drawn through B parallel to CA [Prop. 1.31], and let the 
(straight-line) FH have been drawn through F parallel 
to DE [Prop. 1.31]. Thus, GBCA and DEFH are each 
parallelograms. And GBCA is equal to DEFH. For they 
are on the equal bases BC and EF, and between the 
same parallels BF and GH [Prop. 1.36]. And triangle 
ABC is half of the parallelogram GBCA. For the diago- 
nal AB cuts the latter in half [Prop. 1.34]. And triangle 
FED (is) half of parallelogram DEFH. For the diagonal 
DF cuts the latter in half. [And the halves of equal things 
are equal to one another.] Thus, triangle ABC is equal 
to triangle DEF. 

Thus, triangles which are on equal bases and between 
the same parallels are equal to one another. (Which is) 
the very thing it was required to show. 

Proposition 39 

Equal triangles which are on the same base, and on 
the same side, are also between the same parallels. 

Let ABC and DBC be equal triangles which are on 
the same base BC, and on the same side (of it). I say that 



39 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



auxaTc TtapaXXrjXou; eaxiv. 




Tke^eux'dto yap f) AA- Xeyw, 5xi TtapdXXr)X6c; eaxiv f] 
AA xrj BE. 

Ei yap [Lri, rj/iEko 8ia xoO A ar^eiou xfj BE eCWteia 
7iapdXX/]Xo<; f] AE, xotl e7ie£eu)edio f) EE. I'oov apa eaxl xo 
ABr xpiytovov ifi EBr xpiytovor era xe yap T ^ auxrjc; 
pdaecoc; eaxiv auxco xfjc; Br xotl ev xdig auxdu; 7tapaXXf|Xoi<;. 
dXXd xo ABE xG ABr eaxiv laov xotl xo ABE apa xfi EBE 
laov eaxl xo uelCov ifi eXdaaovr ouep eaxiv d86vaxov oux 
apa TiapdXXr)X6<; eaxiv f] AE xfj BE. 6uo[m<; 8rj Sei^o^iev, 
oxi ou8' aXXr] tic; tiXtjv xfj? AA- f] AA apa xfj BE eaxi 
7iapdXXr]Xo<;. 

Ta apa laa xpiywva xa era xfjc; auxfjc pdaeoc; ovxa xal 
era xa auxa ueprj xal ev xotu; auxau; 7iapaXXf|Xoi<; eaxiv oTtep 
e8ei 8eu;ai. 




Tiaxco laa xptyova xa ABE, EAE era latov pdaewv xwv 
Br, TE xal era xd aOxd uepr). Xeyto, oxi xal ev xdic auxalc 
TtapaXXr]Xoi<; eaxiv. 

'Erae^eux'dco yap f] AA- Xeyw, oxi TtapdXXr]X6<; eaxiv f) 
AA xfj BE. 

Ei yap [Lr\, fj)fdw 8id xou A xrj BE TiapdXXr)Xo? f] AZ, 
xal s%eZ,si>x^ f] ZE. laov apa eaxl xo ABr xpiycovov 
xtp ZrE xpiytovor era xe yap lacov pdaetov eiai xtov Br, 
TE xal ev xalc auxau; TtapaXXr]Xou; xau; BE, AZ. dXXd xo 
ABE xpiycovov laov eaxl iu AEE [xpiycovcp] • xal xo AEE 
apa [xpiycovov] laov eaxl ifi ZEE xpiywvw xo ueTC^ov iw 



they are also between the same parallels. 




For let AD have been joined. I say that AD and BC 
are parallel. 

For, if not, let AE have been drawn through point A 
parallel to the straight-line BC [Prop. 1.31], and let EC 
have been joined. Thus, triangle ABC is equal to triangle 
EBC. For it is on the same base as it, BC, and between 
the same parallels [Prop. 1.37]. But ABC is equal to 
DBC. Thus, DBC is also equal to EBC, the greater to 
the lesser. The very thing is impossible. Thus, AE is not 
parallel to BC. Similarly, we can show that neither (is) 
any other (straight-line) than AD. Thus, AD is parallel 
to BC. 

Thus, equal triangles which are on the same base, and 
on the same side, are also between the same parallels. 
(Which is) the very thing it was required to show. 

Proposition 40^ 

Equal triangles which are on equal bases, and on the 
same side, are also between the same parallels. 




Let ABC and CDE be equal triangles on the equal 
bases BC and CE (respectively), and on the same side 
(of BE) . I say that they are also between the same par- 
allels. 

For let AD have been joined. I say that AD is parallel 
to BE. 

For if not, let AF have been drawn through A parallel 
to BE [Prop. 1.31], and let F E have been joined. Thus, 
triangle ABC is equal to triangle FCE. For they are on 
equal bases, BC and CE, and between the same paral- 
lels, BE and AF [Prop. 1.38]. But, triangle ABC is equal 



40 



ETOIXEIftN a'. 



ELEMENTS BOOK 1 



eXdaaovi/ ouep eaxlv dBuvaxov oux apa TtapdXXr)Xo<; f) AZ 
xfj BE. 6(ioiCK 8r) 8eic;o^ev, oxi oi)B' dXXr] xu; TtXrjv xrjc; AA- 
f] AA apa xfj BE eaxi 7tapdXXr]Xo<;. 

Ta apa Taa xpiywva xa era iawv pdaewv ovxa xal ski xa 
auxa \±epr\ xal ev xau; auxdu; TtapaXXrjXou; eaxiv ouep eBei 
8a^ai. 



to [triangle] DCE. Thus, [triangle] DCE is also equal to 
triangle FCE, the greater to the lesser. The very thing is 
impossible. Thus, AF is not parallel to BE. Similarly, we 
can show that neither (is) any other (straight-line) than 
AD. Thus, AD is parallel to BE. 

Thus, equal triangles which are on equal bases, and 
on the same side, are also between the same parallels. 
(Which is) the very thing it was required to show. 



t This whole proposition is regarded by Heiberg as a relatively early interpolation to the original text. 



\ia'. 

'Edv TtapaXXiqXoypa^ov xpiywvtp pdaiv xe £)(/] xf|V 
auxrjv xal ev xal<; auxdu; TtapaXXrjXou; fj, BmXdaiov eaxi 
xo 7iapaXXr)X6ypa^ov xoO xpiytovou. 

A A E 




B r 

riapaXXr)X6Ypa^ov yap xo ABrA xpiywvcp xfi EBr 
pdaiv xe e^ex^ T, n v auxrjv xrjv Br xal ev xau; auxalc 7ta- 
paXXrjXou; eaxco xau; BT, AE- Xeyw, oxi SmXdaiov laxi xo 
ABrA TiapaXXrjXoypa^ov xou BEr xpiywvou. 

Tke^eux'dw yap f] Ar. Taov Sf| eaxi xo ABr xpiycovov 
x65 EBr xpiywvw- era xe yap xrj<; auxrj<; pdae«<; eaxiv 
auxCS xrj<; Br xal sv xau; auxdu; 7iapaXXr|Xou; xau; Br, AE. 
dXXd xo ABrA TtapaXXr)X6ypa^ov BiTtXdaiov "taxi xou ABr 
xpiyovou- f] yap Ar Sid^iexpo<; auxo Bixa xe^ver &axe 
xo ABrA 7iapaXX/]X6ypa[i^ov xal xou EBr xpiywvou laxl 
BiTtXdaiov. 

'Edv apa KapaXXrjXoypa^iov xpiywvo pdaiv xe exTl T1 1 v 
auxr]v xal ev xau; auxdu; TtapaXXrjXou; fj, BiTtXdaiov eaxi xo 
7iapaXX/]X6ypa[i^ov xou xpiywvou- ouep eBei 8eTc;ai. 

Tco BoiDevxi xpiywvw Taov TtapaXXr]X6ypa^ov auaxf]- 
aaaiSai sv xfj So'deiar) ywvia euiDuypd^^w. 

"Eaxco xo (lev BoiSev xpiycovov xo ABr, f) Be BoiDeToa 
ywvia eu-duypa^oi; rj A- Bel Br] xw ABT xpiywvw laov ua- 
paXXr)X6ypa^ov auaxrjaaa'dai ev xfj A ywvia eu-duypd^w. 



Proposition 41 

If a parallelogram has the same base as a triangle, and 
is between the same parallels, then the parallelogram is 
double (the area) of the triangle. 

A D E 




B C 

For let parallelogram ABCD have the same base BC 
as triangle EBC, and let it be between the same parallels, 
BC and AE. I say that parallelogram ABCD is double 
(the area) of triangle BEC. 

For let AC have been joined. So triangle ABC is equal 
to triangle EBC. For it is on the same base, BC, as 
(EBC), and between the same parallels, BC and AE 
[Prop. 1.37]. But, parallelogram ABCD is double (the 
area) of triangle ABC. For the diagonal AC cuts the for- 
mer in half [Prop. 1.34]. So parallelogram ABCD is also 
double (the area) of triangle EBC. 

Thus, if a parallelogram has the same base as a trian- 
gle, and is between the same parallels, then the parallel- 
ogram is double (the area) of the triangle. (Which is) the 
very thing it was required to show. 

Proposition 42 

To construct a parallelogram equal to a given triangle 
in a given rectilinear angle. 

Let ABC be the given triangle, and D the given recti- 
linear angle. So it is required to construct a parallelogram 
equal to triangle ABC in the rectilinear angle D. 



41 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 




BET 

Tex^a'dco f] Br Bixa xaxd to E, xal era^eux'&co f] AE, 
xal auveaxdxo Ttpoc xfj Er euTJeia xal x£> npbc, auxrj ar^elw 
xo E xrj A ywvia Tar] f] utio TEZ, xal 81a (lev xou A xrj Er 
7iapdXXr]Xo<; ^X'^ co ^ AH, 81a Be xoO T xfj EZ TiapdXXr]Xo<; 
f]xi}co f] TH- rcapaXXrjXoypa^ov dpa eaxl xo ZErH. xal end 
Xar\ eaxlv f) BE xrj Er, Taov eaxl xal xo ABE xptyovov tu 
AEr xpiywvw- era xe yap Tacov pdaecov elai x«v BE, Er xal 
ev xal? auxdig TiapaXXr|Xoi<; xau; Br, AH- BiuXdaiov dpa eaxl 
xo ABr xpiywvov xou AEr xpiyovou. eaxi Be xal xo ZErH 
7iapaXXr)X6ypa[j.(jiov BiTtXdaiov xou AEr xpiywvou- pdaiv xe 
yap auxw xrjv auxrjv exei xal ev xau; auxau; eaxiv auxG 
TiapaXX^XoK;' I'aov dpa eaxl xo ZErH 7tapaXXr]X6ypa^ov 
iw ABr xpiywvw. xal e^ei xr]v uuo TEZ ycoviav Tarjv xrj 
Bo'deiar] xfj A. 

TO dpa Bo'devxi xpiywvw xw ABr Taov TiapaXXrjXoypa^t- 
\±ov auveaxaxai xo ZErH ev ywvia xrj Otto TEZ, f\ut; eaxlv 
Tar] xrj A- oitep eBei rcoirjaai. 



navxog TiapaXXrjXoypd^^iou xtov rcepl xr]v Bid^iexpov rca- 
paXXrjXoypd^wv xd KapaTiXr]p«^axa I'aa dXXr]Xoi<; eaxlv. 

'Eaxw TtapaXXrjXoypa^ov xo ABrA, Bid^texpoc; Be 
auxou f) Ar, rcepl Be xrjv Ar TtapaXXr)X6ypa^(ia ^xev eaxo 
xd E9, ZH, xd Be Xeyojieva TiapaTiXr]p<^axa xd BK, KA- 
Xeyo, oxi Taov eaxl xo BK TcapaTtXrjpwpc xw KA Tiapa- 
TiXrjpwjiaxi. 

'Excel yap rcapaXXr)X6ypa^6v eaxi xo ABrA, Bid^iexpoc; 
Be auxou f] AT, I'aov eaxl xo ABr xplyuvov x£> ArA 
xpiycovw. TidXiv, ercel TiapaXXrjXoypa^t^ov eaxi xo EG, 
Bid^iexpoc; Be auxou eaxiv f) AK, I'aov eaxl xo AEK xpiywvov 
tu A0K xpiywvw. 81a xd auxd Br] xal xo KZr xp(ya>vov 
xo KHr eaxiv I'aov. euel ouv xo jiev AEK xplyovov xo 
A6K xpiycovco eaxlv I'aov, xo Be KZr xt5 KHr, xo AEK 
xpiycovov ^xexa xou KHr Taov eaxl iu A0K xpiytovcp [icxd 
xou KZr- eaxi Be xal oXov xo ABr xplyovov 0X0 xc5 AAr 
laov Xoitiov dpa xo BK 7iapa7iXr]po^a XoikCS xw KA rcapa- 




B E C 



Let BC have been cut in half at E [Prop. 1.10], and 
let AE have been joined. And let (angle) CEF, equal to 
angle D, have been constructed at the point E on the 
straight-line EC [Prop. 1.23]. And let AG have been 
drawn through A parallel to EC [Prop. 1.31], and let CG 
have been drawn through C parallel to EF [Prop. 1.31]. 
Thus, FECG is a parallelogram. And since BE is equal 
to EC, triangle ABE is also equal to triangle AEC. For 
they are on the equal bases, BE and EC, and between 
the same parallels, BC and AG [Prop. 1.38]. Thus, tri- 
angle ABC is double (the area) of triangle AEC. And 
parallelogram FECG is also double (the area) of triangle 
AEC. For it has the same base as (AEC), and is between 
the same parallels as (AEC) [Prop. 1.41]. Thus, paral- 
lelogram FECG is equal to triangle ABC. (FECG) also 
has the angle CEF equal to the given (angle) D. 

Thus, parallelogram FECG, equal to the given trian- 
gle ABC, has been constructed in the angle CEF, which 
is equal to D. (Which is) the very thing it was required 
to do. 

Proposition 43 

For any parallelogram, the complements of the paral- 
lelograms about the diagonal are equal to one another. 

Let ABCD be a parallelogram, and AC its diagonal. 
And let EH and FG be the parallelograms about AC, and 
BK and KD the so-called complements (about AC). I 
say that the complement BK is equal to the complement 
KD. 

For since ABCD is a parallelogram, and AC its diago- 
nal, triangle ABC is equal to triangle ACD [Prop. 1.34]. 
Again, since EH is a parallelogram, and AK is its diago- 
nal, triangle AEK is equal to triangle AHK [Prop. 1.34]. 
So, for the same (reasons), triangle KFC is also equal to 
(triangle) KGC. Therefore, since triangle AEK is equal 
to triangle AHK, and KFC to KGC, triangle AEK plus 
KGC is equal to triangle AHK plus KFC. And the 
whole triangle ABC is also equal to the whole (triangle) 
ADC. Thus, the remaining complement BK is equal to 



42 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



TiX/)p«|j.aT[ eaxiv igov. 



a e a 




b h r 

IlavToc; dpa TtapaXXrjXoYpd^ou /«p[ou xfiv itepl xf]v 
8id^i£xpov 7iapaXX/]XoYpa[i^wv xa 7tapa7tXr)p«uaxa laa dXXr|- 
Xou; eaxiv OTtep s8ei 8eT^ai. 



the remaining complement KD. 



AH D 




B G C 

Thus, for any parallelogramic figure, the comple- 
ments of the parallelograms about the diagonal are equal 
to one another. (Which is) the very thing it was required 
to show. 



[i5'. 

ilocpd xrjv Bo'deiaocv eu'deiav xfi Sodevxi xpiYtovcp !aov 7ta- 
paXXrjXoYpa^ov TtapapaXeiv ev xfj SoOstor) ytovia eu'duYpd^i- 

[ICd. 




Z E K 





b h 


i — 7f 





A A 



"Eaxco f] ^jtev So-delaa eMeia f] AB, xo Be So'dev xpiywvov 
xo T, f] 8e bo'&eiaa ywvia eu'duYpa^tuoc; f] A- 8eT 5f] Ttapa 
xr]V Bo'deTaav sO'deiav xf]v AB xw Bcedevxi xpiYWvw xw T 
laov TtapaXXrjXoYpa^ov TtapapaXeTv ev Tar] xfj A Y^vta. 

Suveaxdxw x65 F xpiY^va) laov napaXX/jXoYpa^^ov xo 
BEZH ev Y«v[a xfj utio EBH, fj eaxiv Tar] xfj A- xa! xdcrdw 
oaxs en' euiSeiac; elvai xf)v BE xfj AB, xa! 5ir)X$co f] ZH 
era xo 0, xa! 81a xou A oiioxepa x£>v BH, EZ 7tapdXX/]Xo<; 
f^x'dw ^ A9, xa! ETC^euy^M f] 6B. xa! erce! etc; napaXXrjXouc; 
xdc; AG, EZ cu-deTa eveTteaev f) 9Z, ai dpa hub A0Z, 0ZE 
Ycoviai 8uaiv op'dau; eiaiv laai. a! dpa Otto BOH, HZE 
860 bp-Q&v eXdaaovec; e£aiv ai 8s dno eXaaaovwv fj 860 
opijfiv sic, dneipov expaXXojievai au^TUTixouaiv ai 6B, ZE 



Proposition 44 

To apply a parallelogram equal to a given triangle to 
a given straight-line in a given rectilinear angle. 




F E K 




HA L 



Let AB be the given straight-line, C the given trian- 
gle, and D the given rectilinear angle. So it is required to 
apply a parallelogram equal to the given triangle C to the 
given straight-line AB in an angle equal to (angle) D. 

Let the parallelogram BEFG, equal to the triangle C, 
have been constructed in the angle EBG, which is equal 
to D [Prop. 1.42]. And let it have been placed so that 
BE is straight-on to AB) And let FG have been drawn 
through to H, and let AH have been drawn through A 
parallel to either of BG or EF [Prop. 1.31], and let HB 
have been joined. And since the straight-line HF falls 
across the parallels AH and EF, the (sum of the) an- 
gles AHF and HFE is thus equal to two right-angles 



43 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



dpa sx|3aXX6|jievai aujjuteaouvxai. expepXrjadwaav xal aufi- 
TUTixextoaav xaxd to K, xal 8id xou K arpeiou OTtoxepa 
x£5v EA, Z6 TiapdXXr]Xo<; f])(i9« f] KA, xal expepXr)aiL)Gjaav 
al 9A, HB erct xa A, M arj^eTa. TiapaXXr]X6ypa^ov apa 
eaxl to 6AKZ, Bidjiexpoc; 5e auxou f) 8K, Tiepi 8s xf)v 9K 
TiapaXXrjXoypajipia y.ev xa AH, ME, xa 8s Xeyo^teva Tiapa- 
7iXr)pco[iaxa xa AB, BZ' iaov apa eaxl xo AB xfii BZ. dXXa 
xo BZ x£> r xpiycjvco eaxlv iaov xal xo AB apa xw T eaxiv 
'iaov. xal etiel Xat] eaxlv f) utio HBE yiovia xfj utio ABM, 
dXXa f] utio HBE xfj A eaxiv Tar), xal f) utio ABM dpa xfj A 
ycovia eaxiv iar). 

napd xfjv SoiSeTaav dpa eui&elav xfjv AB x« So-devxi 
xpiycovco xco r i'aov TiapaXX/]X6ypa|jijj.ov TiapapepX/]xai xo AB 
ev yiovia xrj utio ABM, f] eaxiv Tar) xfj A- oTiep eSei Tioirjaai. 



t This can be achieved using Props. 1.3, 1.23, and 1.31. 

[IE. 

T£i Bo'devxi eu'duypd^w Iaov TiapaXXr]X6ypa[i^.ov auax- 
rjaaadai ev xfj Boifteiar) ywvia eu'duypd^iuw. 

"Eaxco xo ^tev So-dev eu'duypa^iuov xo ABrA, f] Be 
Bo'deTaa ywvia euiSuypa^uoc; f] E- 5eT §f] xw ABrA eO'du- 
ypd^tp iaov TiapaXXr)X6ypa^iuov auoxf]aaa'f)ai ev xfj Bo-deia?) 
yovia xfj E. 

'ErceCsux'dw f] AB, xal auveaxdxo xw ABA xpiywvw 
iaov TtapaXXr]X6ypa^ov xo Z9 ev xrj utio 9KZ ywvia, f) 
eaxiv Tar] xfj E- xal TiapapepXf]adw Tiapa xf]v H9 euiMav iw 
ABr xpiycivo iaov TiapaXXr]X6ypa^ov xo HM ev xfj utio 
H9M ytovia, f\ eaxiv iar) xfj E. xal CTiei f) E ycovia exaxepa 
xGv utio 9KZ, H9M eaxiv Tar), xal f] utio 9KZ dpa xfj utio 
H9M eaxiv 'iar). xoivf] Tipoaxeia'dw f] utio K9H- ai dpa 
utio ZK9, K9H xdi<; utio K9H, H9M i'aai eiaiv. dXX' ai 
utio ZK9, K9H 8ualv opiate; Taai eiaiv xal ai utio K9H, 
H9M dpa Buo op'dau; Taai eiaiv. Tip6<; Sf] xivi eu'dela xfj H9 
xal tu Tipoc auxfj ar^eio x<3 9 Buo euiSeTai ai K9, 9M [lt] 
era xa auxd [ispf] xei^ievai xa<; ecpe^fj<; y«via<; Buo opiJau; 
Taa<; Tioiouaiv tn euiJeiai; apa eaxlv f] K9 xfj 9M- xal 
cticI tic, TiapaXXf]Xou<; xd<; KM, ZH euiJeTa eveiieaev f) 9H, 
ai evaXXac; ycoviai ai utio M9H, 9HZ Taai dXXf|Xaic eiaiv. 
xoivf) Tipoaxeia'dcL) f] utio 9HA- ai dpa utio M9H, 9HA xau; 
utio 9HZ, 9 HA i'aai eiaiv. dXX' ai utio M9H, 9HA 8uo 
op'&dic, laai eiaiv xal ai utio 9HZ, 9HA apa Suo op^dit; 
laai eiaiv cti' euiDeiat; dpa eaxiv f) ZH xfj HA. xal cticI f) 
ZK xfj 9H Xar] xe xal TiapdXXr]X6<; eaxiv, dXXa xal f] 9H xfj 
MA, xal f) KZ apa xfj MA iar] xe xal TiapdXXr)X6<; eaxiv xal 



[Prop. 1.29]. Thus, (the sum of) BHG and GFE is less 
than two right-angles. And (straight-lines) produced to 
infinity from (internal angles whose sum is) less than two 
right-angles meet together [Post. 5]. Thus, being pro- 
duced, HB and FE will meet together. Let them have 
been produced, and let them meet together at K. And let 
KL have been drawn through point K parallel to either 
of EA or FH [Prop. 1.31]. And let HA and GB have 
been produced to points L and M (respectively) . Thus, 
HLKF is a parallelogram, and H K its diagonal. And 
AG and ME (are) parallelograms, and LB and BF the 
so-called complements, about HK. Thus, LB is equal to 
BF [Prop. 1.43]. But, BF is equal to triangle C. Thus, 
LB is also equal to C. Also, since angle GBE is equal to 
ABM [Prop. 1.15], but GBE is equal to D, ABM is thus 
also equal to angle D. 

Thus, the parallelogram LB, equal to the given trian- 
gle C, has been applied to the given straight-line AB in 
the angle ABM, which is equal to D. (Which is) the very 
thing it was required to do. 



Proposition 45 

To construct a parallelogram equal to a given rectilin- 
ear figure in a given rectilinear angle. 

Let ABCD be the given rectilinear figure^ and E the 
given rectilinear angle. So it is required to construct a 
parallelogram equal to the rectilinear figure ABCD in 
the given angle E. 

Let DB have been joined, and let the parallelogram 
FH, equal to the triangle ABD, have been constructed 
in the angle HKF, which is equal to E [Prop. 1.42]. And 
let the parallelogram GM, equal to the triangle DBG, 
have been applied to the straight-line GH in the angle 
GHM, which is equal to E [Prop. 1.44]. And since angle 
E is equal to each of (angles) HKF and GHM, (an- 
gle) HKF is thus also equal to GHM. Let KHG have 
been added to both. Thus, (the sum of) FKH and KHG 
is equal to (the sum of) KHG and GHM. But, (the 
sum of) FKH and KHG is equal to two right-angles 
[Prop. 1.29]. Thus, (the sum of) KHG and GHM is 
also equal to two right-angles. So two straight-lines, KH 
and HM, not lying on the same side, make adjacent an- 
gles with some straight-line GH, at the point H on it, 
(whose sum is) equal to two right-angles. Thus, KH is 
straight-on to HM [Prop. 1.14]. And since the straight- 
line HG falls across the parallels KM and FG, the al- 
ternate angles MHG and HGF are equal to one another 
[Prop. 1.29]. Let HGL have been added to both. Thus, 
(the sum of) MHG and HGL is equal to (the sum of) 



44 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



era^suyvuoucuv auTa? eMelai ai KM, ZA- xai al KM, ZA 
apa I'aai te xal 7iapdXX/]Xo( elaiv 7tapaXX/]X6ypa^|iov dpa 
ecm to KZAM. xal end loov eoxl to ^iev ABA xp[y«vov 
to Z0 7iapaXX/)Xoypd|ji[j.cp, to 8s ABr tG HM, 6Xov dpa 
to ABrA eu^uypa^ov 6Xcp tc3 KZAM TTapaXXrjXoypdjUjiw 
screw iaov. 



A 




k e m 

Tco apa BoiDevTi Eu , duypd|i^.cp tw ABrA loov uapaX- 
Xr]X6ypa^ov auveoTaxai to KZAM ev yovia xfj Otio ZKM, 
f) eaxiv Tot) Tfj So'deior] xfj E- ouep e8si Tioifjaai. 

t The proof is only given for a four-sided figure. However, the extension 

Ako Tfj? 8o , de(or]<; su-delac. xerxpdywvov dvaypdtjjai. 

'Eoxo #] Bo-fMaa euTMa f] AB- 8eT 5f| duo xrjc AB 
eO'deiac; xexpdycovov dvaypdtjiai. 

"Hy^co xfj AB euiiteia diio toO Tipoc. auxfj ar)[ie(ou xou 
A itpoc. opM? f\ AT, xal xeicrdco Tfj AB lot] f\ AA- xal 8ia 
^iev tou A ar)[ie(ou xfj AB 7iapdXXr]Xoc. fix^to f] AE, 8ia 
8e toO B ar)[i£Lou Tfj AA 7tapdXXr]Xo<; rjx$co f] BE. napaX- 
X/]X6ypa[i^ov apa eaxl to AAEB- tar] apa soxlv f] ^isv AB 
Tfj AE, f] 8e AA Tfj BE. dXXa f] AB Tfj AA eaxiv larj- 
ai xeaaape? dpa ai BA, AA, AE, EB '{aai dXXf]Xaic sicnv 
iaoTtXsupov apa ecru to AAEB TtapaXXr)X6ypa[i(jiov. Xeyco 
8r], otl xal opiJoyoviov. enel yap el? uapaXX^Xou? xd? AB, 
AE su'dela eveueaev f\ AA, al dpa (mo BAA, AAE ycovtai 
8uo op-ddl? '(aai eialv. dpfty) 8e f] Otio BAA- op'df] dpa xal 



HGF and HGL. But, (the sum of) MHG and ffGL is 
equal to two right-angles [Prop. 1.29]. Thus, (the sum of) 
HGF and i/GL is also equal to two right-angles. Thus, 
FG is straight-on to GL [Prop. 1.14]. And since FK is 
equal and parallel to EG [Prop. 1.34], but also HG to 
ML [Prop. 1.34], KF is thus also equal and parallel to 
ML [Prop. 1.30]. And the straight-lines KM and FL 
join them. Thus, KM and FL are equal and parallel as 
well [Prop. 1.33]. Thus, KFLM is a parallelogram. And 
since triangle ABD is equal to parallelogram FH, and 
DBC to GM, the whole rectilinear figure ABCD is thus 
equal to the whole parallelogram KFLM. 



D 




K H M 



Thus, the parallelogram KFLM, equal to the given 
rectilinear figure ABCD, has been constructed in the an- 
gle FKM, which is equal to the given (angle) E. (Which 
is) the very thing it was required to do. 

to many-sided figures is trivial. 

Proposition 46 

To describe a square on a given straight-line. 

Let AB be the given straight-line. So it is required to 
describe a square on the straight-line AB. 

Let AC have been drawn at right-angles to the 
straight-line AB from the point A on it [Prop. 1.11], 
and let AD have been made equal to AB [Prop. 1.3]. 
And let DE have been drawn through point D parallel to 
AB [Prop. 1.31], and let BE have been drawn through 
point B parallel to AD [Prop. 1.31]. Thus, ADEB is a 
parallelogram. Therefore, AB is equal to DE, and AD to 
BE [Prop. 1.34]. But, AB is equal to AD. Thus, the four 
(sides) BA, AD, DE, and EB are equal to one another. 
Thus, the parallelogram ADEB is equilateral. So I say 
that (it is) also right-angled. For since the straight-line 



45 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



f) Otio AAE. twv 8e 7iapaXX/]Xoypd|jijj.(x>v )(«p[wv ai dme- 
vavxiov TtXeupai xe xal ywviai i'aai dXXfjXaic; eiaiv op-df) dpa 
xai exaxepa xc5v dnevavxiov xGv utio ABE, BEA ycovicov 
op^oycoviov apa eaxl to AAEB. eSeix'dr] 8e xal iaoftXeupov. 



r 



A E 



A B 

Texpdywvov apa eaxiv xa[ eaxiv arco xfjc; AB eu-deiac; 
dvayeypa^ievov onep eSei Troifjaai. 

'Ev xolc; op^oyMvioic; xpiycovoig xo duo xfjc; xfjv op^rjv 
ycoviav un;oxeivouar]<; TtXeupac; xexpdywvov laov eaxl xou; 
duo iwv xf]V 6p$r]v ycoviav Ttepiexouawv TtXeupwv xe- 
xpaywvou;. 

"Eaxto xpiywvov op-doycoviov xo ABr op^/jv exov xrjv 
uno BAr ycoviav Xeyo, oxi xo duo xfjc; Br xexpdycovov 
Taov eaxl xolc dfto xcov BA, Ar xexpaywvoic;. 

Avayeypdcp'dw yap and y.ev xfjc; Br xexpdyiovov xo 
BAEr, duo 8e xwv BA, Ar xd HB, 9r, xai Bid xoO 
A oTtoxepa xwv BA, TE 7tapdXXr]Xoc; fix^" f) AA- xal 
ETieCeux'dwaav ai AA, ZT. xal ind 6pi5f| eaxiv exaxepa 
iwv utio BAr, BAH ywviwv, Ttpoc; 8f] xivi cu-Ma xfj BA 
xal x£S Ttpoc; auxfj arjueiM ifi A 860 eui9eTai ai Ar, AH ^tf) 
excl xd auxd \iepf) xei^ievai xac; ecpec;fjc; ycoviac; 8ualv opiJaTc; 
Taac; TtoioOaiv z% eO'deiac; apa eaxiv f] FA xfj AH. 81a xd 
auxd 8f) xal f) BA xfj A0 eaxiv tn eO'deiac;. xal enel tar] 
eaxlv f] O716 ABr ywvia xfj uko ZBA- opiSf] yap exaxepa- 
xotvf) npoaxeia'dw f] tmo ABr- oXrj apa f\ hub ABA oXr] xfj 
O716 ZBr eaxiv lay), xal ckcI iar) eaxlv f) ^xev AB xfj Br, f) 
8e ZB xfj BA, 860 8f] ai AB, BA 860 xdic; ZB, Br laai eiaiv 
exaxepa exaxepa- xai ywvia f] 6716 ABA yovia xfj bub ZBr 
Tar) - pdaic; apa f] AA pdaei xfj Zr [eaxiv] Iar], xal xo ABA 



AD falls across the parallels AB and DE, the (sum of 
the) angles BAD and ADE is equal to two right-angles 
[Prop. 1.29]. But BAD (is a) right-angle. Thus, ADE 
(is) also a right-angle. And for parallelogrammic figures, 
the opposite sides and angles are equal to one another 
[Prop. 1.34]. Thus, each of the opposite angles ABE 
and BED (are) also right-angles. Thus, ADEB is right- 
angled. And it was also shown (to be) equilateral. 

c 



D E 



A B 

Thus, {ADEB) is a square [Def. 1.22]. And it is de- 
scribed on the straight-line AB. (Which is) the very thing 
it was required to do. 

Proposition 47 

In right-angled triangles, the square on the side sub- 
tending the right-angle is equal to the (sum of the) 
squares on the sides containing the right-angle. 

Let ABC be a right-angled triangle having the angle 
BACa right-angle. I say that the square on BC is equal 
to the (sum of the) squares on BA and AC. 

For let the square BDEC have been described on 
BC, and (the squares) GB and HC on AB and AC 
(respectively) [Prop. 1.46]. And let AL have been 
drawn through point A parallel to either of BD or CE 
[Prop. 1.31]. And let AD and FC have been joined. And 
since angles BAG and BAG are each right-angles, then 
two straight-lines AC and AG, not lying on the same 
side, make the adjacent angles with some straight-line 
BA, at the point A on it, (whose sum is) equal to two 
right-angles. Thus, CA is straight-on to AG [Prop. 1.14]. 
So, for the same (reasons), BA is also straight-on to AH. 
And since angle DBC is equal to FBA, for (they are) 
both right-angles, let ABC have been added to both. 
Thus, the whole (angle) DBA is equal to the whole (an- 
gle) FBC. And since DB is equal to BC, and FB to 
BA, the two (straight-lines) DB, BA are equal to the 



4G 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



xpiywvov xG ZBr xpiywvw saxlv iaov xa( [eaxi] xou ^.ev 
ABA xpiyovou 8iitXdaiov xo BA TiapaXXr)X6ypa^i|iov pdaiv 
xe yap xfjv auxfjv s^ouai xf]v BA xal ev xau; auxau; elai 
7tapaXXf]Xoi<; xau; BA, AA - xou Se ZBr xpiywvou 8i7tXdaiov 
xo HB xexpdywvov pdaiv xe yap ndXiv xf]v auxrjv e)(ouai 
xf]v ZB xal ev xau; auxau; eiai TtapaXXf|Xou; xau; ZB, Hr. 
[xa 8e xwv iaov 8iTiXdaia laa dXXf|Xou; eaxiv] iaov apa eaxi 
xal xo BA KapaXXrjXoypa^^iov xw HB xexpaywvw. 6[ioiok 
8f] eTuCeuyvu^evwv xwv AE, BK Beix^rpexai xal xo TA 
7tapaXX/]X6ypa^ov I'aov iw 0T xexpaywvy oXov apa xo 
BAEr xexpdywvov 8uol xou; HB, 8r xexpaywvou; iaov 
eaxiv. xal eaxi xo ^iev BAEr xexpdycovov drco xfjc Br dva- 
ypacpev, xa 8e HB, OT and xcov BA, AI\ xo apa anb xfjc; 
Br TtXeupac; xexpdyovov iaov eaxi xolc duo xfiv BA, Ar 
uXeupov xexpayovou;. 




A A E 
'Ev apa xou; op'doyovioic; xpiywvoic; xo dno xfjc; xfjv 
op'dfjv ywviav UTtoxeivouarjc; TtXeupdc; xexpdyiovov i'aov eaxi 
xoTc; duo xGv xfjv 6pi3f)v [ywviav] rcepiexouacov TtXeupcov xe- 
xpaywvoic oTtep eSei Sel^ai. 

t The Greek text has "FB, BC", which is obviously a mistake, 
t This is an additional common notion. 



two (straight-lines) CB, BF,^ respectively. And angle 
DBA (is) equal to angle FBC. Thus, the base AD [is] 
equal to the base FC, and the triangle ABD is equal to 
the triangle FBC [Prop. 1.4]. And parallelogram BL 
[is] double (the area) of triangle ABD. For they have 
the same base, BD, and are between the same parallels, 
BD and AL [Prop. 1.41]. And square GB is double (the 
area) of triangle FBC. For again they have the same 
base, FB, and are between the same parallels, FB and 
GC [Prop. 1.41]. [And the doubles of equal things are 
equal to one another.]- 1 - Thus, the parallelogram BL is 
also equal to the square GB. So, similarly, AE and BK 
being joined, the parallelogram CL can be shown (to 
be) equal to the square HC. Thus, the whole square 
BDEC is equal to the (sum of the) two squares GB and 
HC. And the square BDEC is described on BC, and 
the (squares) GB and HC on BA and AC (respectively) . 
Thus, the square on the side BC is equal to the (sum of 
the) squares on the sides BA and AC. 

H 




D L E 
Thus, in right-angled triangles, the square on the 
side subtending the right-angle is equal to the (sum of 
the) squares on the sides surrounding the right- [angle]. 
(Which is) the very thing it was required to show. 



47 



ETOIXEIfiN a'. 



ELEMENTS BOOK 1 



'Edv xpiycovou to and [iiag xfiv TiXeupfiv xexpdycovov 
laov fj xolc; duo xfiv Xoittcov xou xpiycovou 860 TtXeupcov 
xexpaycovou;, f) Tt£pi£)(o|jievr] ycovia u^o Xomcov xou 
xpiycovou 860 TtXeupcov opfir] saxiv. 




A A B 

Tpiycovou yap xou ABr xo dno [iiac, xfjg Br TtXeupac; 
xexpdycovov l'aov saxco xou; dno xcov BA, Ar TtXeupcov xe- 
xpaycovou;- Xeyco, 6x1 op'dr] eaxiv f) utio BAr ycovia. 

"H)edco yap diio xou A ar\\ieiou xrj Ar eu'deia Tipo<; 6pi5d<; 
f] A A xal xeiaiJco xfj BA Tar] f) AA, xal eTieCeuyiJco f] Ar. 
End Tar] eaxlv f) A A xfj AB, laov eaxl xal xo duo xrjc 
AA xexpdycovov xo duo xfj? AB xexpaycovcp. xoivov Tipo- 
axeiai9co xo and xrjc; Ar xexpdycovov xd dpa duo xcov AA, 
Ar xexpdycova Taa eaxl xou; arco xcov BA, Ar xexpaycovou;. 
dXXd xou; [iev and xcov AA, Ar Taov eaxl xo ano xrjg AE 
6pTL>r) yap eaxiv f) utio AAr ycovia- xou; 8s duo xcov BA, 
Ar laov eaxl xo duo xfjc; Br- UTtoxeixai yap- xo dpa arco 
xrjc; Ar xexpdycovov laov sax! xco dno xrjc; Br xexpaycovcp- 
coaxe xal TtXeupa f) Ar xrj Br eaxiv Iar]- xal end Tar) eaxlv 
f) AA xfj AB, xoivr) 8s f] AT, Suo 8/) ai AA, Ar 660 xdu; 
BA, Ar Taai eiaiv xal pdaic; f) Ar pdaei xfj Br lot)- ycovia 
dpa f) utio AAr ycovia xrj utio BAr [eaxiv] Iar). 6pi5r] 8e f] 
utio AAr- op-Qr) dpa xal f] utio BAr. 

'Edv dpa xpiycovou xo duo [ii&c, xcov TtXeupcov xexpdycovov 
laov fj xolc; duo xcov Xoiticov xou xpiycovou 860 TtXeupcov 
xexpaycovou;, f) Ttepie/o|jievr] ycovia utio xcov Xoiticov xou 
xpiycovou 860 TtXeupcov op-dr] eaxiv oTtep eSei SeT^ai. 



Proposition 48 

If the square on one of the sides of a triangle is equal 
to the (sum of the) squares on the two remaining sides of 
the triangle then the angle contained by the two remain- 
ing sides of the triangle is a right-angle. 



C 




DAB 



For let the square on one of the sides, BC, of triangle 
ABC be equal to the (sum of the) squares on the sides 
BA and AC. I say that angle BAC is a right-angle. 

For let AD have been drawn from point A at right- 
angles to the straight-line AC [Prop. 1.11], and let AD 
have been made equal to BA [Prop. 1.3], and let DC 
have been joined. Since DA is equal to AB, the square 
on DA is thus also equal to the square on AB) Let the 
square on AC have been added to both. Thus, the (sum 
of the) squares on DA and AC is equal to the (sum of 
the) squares on BA and AC. But, the (square) on DC is 
equal to the (sum of the squares) on DA and AC. For an- 
gle DAC is a right-angle [Prop. 1.47]. But, the (square) 
on BC is equal to (sum of the squares) on BA and AC. 
For (that) was assumed. Thus, the square on DC is equal 
to the square on BC. So side DC is also equal to (side) 
BC. And since DA is equal to AB, and AC (is) com- 
mon, the two (straight-lines) DA, AC are equal to the 
two (straight-lines) BA, AC. And the base DC is equal 
to the base BC. Thus, angle DAC [is] equal to angle 
BAC [Prop. 1.8]. But DAC is a right-angle. Thus, BAC 
is also a right-angle. 

Thus, if the square on one of the sides of a triangle is 
equal to the (sum of the) squares on the remaining two 
sides of the triangle then the angle contained by the re- 
maining two sides of the triangle is a right-angle. (Which 
is) the very thing it was required to show. 



t Here, use is made of the additional common notion that the squares of equal things are themselves equal. Later on, the inverse notion is used. 



48 



ELEMENTS BOOK 2 

Fundamentals of Geometric Algebra 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



"Opoi. 

a'. Ilav TTapaXXr)X6ypa[j.(jiov opi&oywviov Tiepiexea-dai 
Xeyexai utio Buo xfiv xf]v opiS/jv yiovlav TiepiexouaCiv 
eMeifiv. 

P'. IlavToc; 8s TiapaXX/]Xoypd[ji^ou x^plou iSv rcepl xf)v 
8idpiexpov auxoO TiapaXXrjXoypd^ojv ev ottoiovouv auv xoT<; 
5ua! TiapaTiXr)p«|jiaai yvw^wv xaXeladio. 



Definitions 

1. Any rectangular parallelogram is said to be con- 
tained by the two straight-lines containing the right- 
angle. 

2. And in any parallelogrammic figure, let any one 
whatsoever of the parallelograms about its diagonal, 
(taken) with its two complements, be called a gnomon. 



'Edv Sai 8uo eOdelai, Tji.rj'dfj 8e f] exepa auxwv etc; oaa- 
8r)7ioxoOv x^ir][iaxa, xo Tiepiexojievov op^oycoviov utio xwv 
8uo eO'dei&v laov eaxl xou; utio xe xrjc; dx[if|xou xod exdaxou 
x£>v x^ir^dxtov Tiepiexo[ievoi<; op-doycovioic;. 



A- 



B 











K 


A 





H 

Z 

TEaxoaav 8uo einMai ai A, Br, xal xex^o'dw f) Br, 
cbc; exuxev, xaxd xd A, E ar)^eTa- Xeyw, oxi xo utio twv A, 
Br Tiepiexo|ievov opiSoytoviov laov eaxl xco xe utio xc5v A, 
BA Tiepiexo^ievtp op^oywviw xal xw utio xwv A, AE xal exi 
xfi utio x«v A, EE 

"H)cdco yap and xou B xfj Br Tipog opiJdc; f) BZ, xal 
xeladio xfj A for) f] BH, xal Sid [Lev xou H xfj Br TiapdXXr]Xo<; 
rjX'dw f] H0, 8id 8e xwv A, E, T xfj BH TiapdXX/]Xoi fjx'dwaav 
aiAK, EA, TO. 

'laov 8f| eaxi xo B9 xoI<; BK, AA, E0. xal eaxi xo 
^iev B6 xo utio xwv A, Br- Tiepiexexai jiev yap utio xwv 
HB, Br, I'ar) 8e f\ BH xfj A- xo 8e BK xo utio xwv A, BA- 
Tiepiexexai ^xev yap utio xwv HB, BA, iar) 8e f] BH xfj A. xo 
8e AA xo utio xwv A, AE- i'ar) yap f) AK, xouxeaxiv f] BH, 
xfj A. xal exi ojioiclx; xo E0 xo utio xwv A, EE xo apa utio 
x£Sv A, Br laov eaxl xw xe utio A, BA xal xo utio A, AE 
xal exi tu utio A, EE 

'Edv apa Sai 8uo eu'delai, xpydfj Be f] exepa auxGv tic, 
6aa8/]Tioxouv x[if]^axa, xo Tiepiexopievov op'doyoviov utio 
xwv Buo eu-deiwv I'aov eaxl xou; utio xe xfj<; dx^xou xal 
exdaxou xGv x[i/]^dxwv Tiepiexo^evou; op-doyMvlou;- oTiep 



Proposition 1+ 

If there are two straight-lines, and one of them is cut 
into any number of pieces whatsoever, then the rectangle 
contained by the two straight-lines is equal to the (sum 
of the) rectangles contained by the uncut (straight-line), 
and every one of the pieces (of the cut straight-line) . 

A 



B 



D 



K 



H 



Let A and BC be the two straight-lines, and let BC 
be cut, at random, at points D and E. I say that the rect- 
angle contained by A and BC is equal to the rectangle(s) 
contained by A and BD, by A and DE, and, finally, by A 
and EC. 

For let BF have been drawn from point B, at right- 
angles to BC [Prop. 1.11], and let BG be made equal 
to A [Prop. 1.3], and let GH have been drawn through 
(point) G, parallel to BC [Prop. 1.31], and let DK, EL, 
and CH have been drawn through (points) D, E, and C 
(respectively), parallel to BG [Prop. 1.31]. 

So the (rectangle) BH is equal to the (rectangles) 
BK, DL, and EH. And BH is the (rectangle contained) 
by A and BC. For it is contained by GB and BC, and BG 
(is) equal to A. And BK (is) the (rectangle contained) by 
A and BD. For it is contained by GB and BD, and BG 
(is) equal to A. And DL (is) the (rectangle contained) by 
A and DE. For DK, that is to say BG [Prop. 1.34], (is) 
equal to A. Similarly, EH (is) also the (rectangle con- 
tained) by A and EC. Thus, the (rectangle contained) 
by A and BC is equal to the (rectangles contained) by A 



50 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



e8ei 8eT<;ou. 



and BD, by A and DE, and, finally, by A and EC. 

Thus, if there are two straight-lines, and one of them 
is cut into any number of pieces whatsoever, then the 
rectangle contained by the two straight-lines is equal 
to the (sum of the) rectangles contained by the uncut 
(straight-line), and every one of the pieces (of the cut 
straight-line) . (Which is) the very thing it was required 
to show. 



t This proposition is a geometric version of the algebraic identity: a(b + c + c2 H ) = ab + ac + ad + ■ 



P'- 

'Edv eO'dsTa ypa^iuf] xur^fj, cdc, exu)(ev, to Otto xfjc; 6kr\<z 
xal exaxepou xGv xur^dxcov Ttepie)(6pievov opiJoycoviov iaov 
eaxl xo arco xfjc; oArjc; xexpayovcp. 



a r 



B 



Proposition 2+ 

If a straight-line is cut at random then the (sum of 
the) rectangle (s) contained by the whole (straight-line), 
and each of the pieces (of the straight-line), is equal to 
the square on the whole. 

AC B 



A Z E 

Eu'deTa yap f) AB xex[ifjad<j, ok exu)(ev, xaxa xo T 
o/)[ieTov Xeyo, oxi xo utco xwv AB, Br Tcepie)(6^evov 
op^oyMviov jiexd xou utco BA, Ar Tcepie)(o|jievou op-do- 
yoviou laov eaxl xw duo xfjc; AB xexpaytovio. 

Avayeypdcp'do yap diro xfjc; AB xexpdytovov xo AAEB, 
xal fix&oi Sid xou T oTcoxepa xwv AA, BE TiapdXXrjXoc; f) 

rz. 

'laov Sfj eaxl xo AE xou; AZ, TE. xa( eaxi xo ^.ev AE 
xo dico xfjc; AB xexpdywvov, xo 8s AZ xo Otco xGv BA, 
Ar Tiepie)(o^i£vov opTSoytoviov Tcepie)(exai ydp otco xwv 
A A, Ar, lar\ 8e f) A A xfj AB- xo 8e TE xo Otco xfiv AB, 
Br- iar) yap f) BE xfj AB. xo dpa utco xfiv BA, Ar jiexa 
xou utco xtov AB, Br laov eaxl xw aTco xfjc; AB xexpayova). 

'Eav dpa eu'de'ia ypa^f) x^trydfj, wc; exu)(ev, xo utco xfjc; 
oX/]c; xal exaxepou xwv x^irj^dxwv Tcepie)(6^evov op'doywvi.ov 
laov eaxl x£5 aTco xfjc; oXrjc; xexpaywvw- OTcep e8ei 8eu;ai. 



D F E 

For let the straight-line AB have been cut, at random, 
at point C. I say that the rectangle contained by AB and 
BC, plus the rectangle contained by BA and AC, is equal 
to the square on AB. 

For let the square ADEB have been described on AB 
[Prop. 1.46], and let CF have been drawn through C, 
parallel to either of AD or BE [Prop. 1.31]. 

So the (square) AE is equal to the (rectangles) AF 
and CE. And AE is the square on AB. And AF (is) the 
rectangle contained by the (straight-lines) BA and AC. 
For it is contained by DA and AC, and AD (is) equal to 
AB. And CE (is) the (rectangle contained) by AB and 
BC. For BE (is) equal to AB. Thus, the (rectangle con- 
tained) by BA and AC, plus the (rectangle contained) by 
AB and BC, is equal to the square on AB. 

Thus, if a straight-line is cut at random then the (sum 
of the) rectangle (s) contained by the whole (straight- 
line), and each of the pieces (of the straight-line), is equal 
to the square on the whole. (Which is) the very thing it 
was required to show. 



51 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



t This proposition is a geometric version of the algebraic identity: a b + 

Y • 

'Edv eO'dsTa ypa[iur] T^y^df), cbc, exuxev, to utco xrjc oXrjc 
xal evoc, xGv xjir^dxtov Tcepiexo^ievov opiDoyoviov iaov sail 
to xe utco tov x^ir)^dxcov Tcepiexo^iivw opTJoyMvltp xal iS 
aTco toD Tcpoeipr)^evou xfjirinaxoc, xexpaytovw. 

a r b 



Z A E 

EO'deTa yap f] AB xex^yja'dto, <l>c exuxev, xaxd to E 
Xeyco, oxi to utco twv AB, Br Tcepiexo^ievov opiSoycoviov 
iaov eaxl xfi xe Otto twv Ar, TB Tcepiexo^ievw opiJoyovia) 
^.exd tou dico xrjc, Br xexpaywvou. 

AvayeypdcpTJco Y^P aTI ° -TB xexpdywvov to TAEB, 
xal BirjX'dw ttj EA em to Z, xal Sid tou A oTcoxepa xcov FA, 
BE TcapdXXr)Xoc, f)x&M f\ AZ. iaov Sr] eaxi to AE tou; AA, 
TE' xal 6cm to [lzv AE to utco xtov AB, Br Tcepiex6[ievov 
opiDoycoviov Tcepiexexai [Lev yap utco iuv AB, BE, iar\ Se f] 
BE xfj Br- to 8s AA to utco tuv Ar, TB- lot] yap f\ AT 
xfj TB' to 8e AB to duo xfjc TB xexpaycovov xo apa utco 
xwv AB, Br Tcepiexo^tevov opiDoycoviov laov eaxl ifi utco 
icov Ar, TB Tcepiexo^tevw op-doycovlo jiexd tou aTco xfj? Br 
xexpaytovou. 

'Eav apa eu-dela ypajji^tr] T[a.rj'df), cbc; exuxev, xo utco xrjc, 
oXr]C. xal evoc. xfiv x^irj^dxwv Tcepiexojievov opiJoycoviov laov 
eaxl xO xe utco xtov xjir^dxwv Tcepiexo^ievtL> op-doytovlo xal 
x<3 dTco xou Tcpo£ipr)[ievou x^rpaxoc, xexpaycovcp- oTcep eSei 
8eT<;ai. 



t This proposition is a geometric version of the algebraic identity: (a + 

5'. 

'Eav eu'deTa yp°Wn "■W^ exuxev, xo duo xfjc; 
oXr]C_ xexpdywvov laov eaxl xolc xe aTco xwv x^r)[idx«v xe- 
xpaycovoic. xal xQ 81? utco xwv x^tr]udxov Tcepiexo^tevw op'do- 



c = a 2 if a = b + c. 

Proposition 3 1 " 

If a straight-line is cut at random then the rectangle 
contained by the whole (straight-line), and one of the 
pieces (of the straight-line), is equal to the rectangle con- 
tained by (both of) the pieces, and the square on the 
aforementioned piece. 

AC B 



F D E 

For let the straight-line AB have been cut, at random, 
at (point) C. I say that the rectangle contained by AB 
and BC is equal to the rectangle contained by AC and 
CB, plus the square on BC. 

For let the square CDEB have been described on CB 
[Prop. 1.46], and let ED have been drawn through to 
F, and let AF have been drawn through A, parallel to 
either of CD or BE [Prop. 1.31]. So the (rectangle) AE 
is equal to the (rectangle) AD and the (square) CE. And 
AE is the rectangle contained by AB and BC. For it is 
contained by AB and BE, and BE (is) equal to BC. And 
AD (is) the (rectangle contained) by AC and CB. For 
DC (is) equal to CB. And DB (is) the square on CB. 
Thus, the rectangle contained by AB and BC is equal to 
the rectangle contained by AC and CB, plus the square 
on BC. 

Thus, if a straight-line is cut at random then the rect- 
angle contained by the whole (straight-line), and one of 
the pieces (of the straight-line), is equal to the rectangle 
contained by (both of) the pieces, and the square on the 
aforementioned piece. (Which is) the very thing it was 
required to show. 

a = ab + a 2 . 

Proposition 4" 1 " 

If a straight-line is cut at random then the square 
on the whole (straight-line) is equal to the (sum of the) 
squares on the pieces (of the straight-line), and twice the 



52 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



ytovicp. 



A 



r b 



rectangle contained by the pieces. 

A C 



e 









H 



K 



A Z E 

EuiMa yap ypamif] f) AB xex^ifjadco, cbc; exu)(ev, xaxd 
to T. Xeyco, oxi to dira xfjc AB xexpdycovov laov eaxl xoTc 
xe dno xcov Ar, TB xexpaycovoic xai xcp Sic; Otto xcov Ar, 
TB Ttepie/o|jievcp opOoycovico. 

Avayeypdcp'dco yap aTto xfjc AB xexpdycovov xo AAEB, 
xai CTieCeux^co BA, xal 8id ^ev xou T oTioxepa xcov AA, 
EB TtapdXX/jXoc fixiSco ^ TZ, 8id 8e xou H OTioxepa xcov AB, 
AE TiapdXXiqXoc fjy^co f] 0K. xal insl TiapdXXiqXoc eaxiv f] 
TZ xfj AA, xal eic auxdc e^iTieTixcoxev f) BA, f] exxoc ycovia 
f) Otto THB i'ar) eaxl xfj evxoc xal dnevavxbv xfj Otto AAB. 
dXX' f) utto AAB xfj Otto ABA eaxiv lot], snel xal TiXeupd f] 
BA xfj A A eaxiv lar)' xal f) Otto THB dpa ycovid xfj Otto HBr 
eaxiv larj- uoie xal TiXeupd f] BT TtXeupa xfj TH eaxiv iar)- 
dXX' fj [iev TB xfj HK eaxiv for). f\ 5e TH xfj KB- xal f) HK 
dpa xfj KB eaxiv for)' iaoTtXeupov dpa eaxl xo THKB. Xeyco 
Br), oxi xal 6pi!)oywviov. etcei yap TiapdXXr)X6c eaxiv f] TH 
xfj BK [xal eic auxdc e^TieTcxcoxev eu$eTa rj TB] , ai dpa Otto 
KBr, HrB ycoviai 8uo op'daic eiaiv iaai. opiDf] 8e f] Otco 
KBE op'df] dpa xal f] Otio BITE coaxe xal ai dnevavxiov 
ai Otto THK, HKB opiJai eiaiv. op-doycoviov dpa eaxl xo 
THKB' e8eix'dT) Be xal iaouXeupov xexpdycovov dpa eaxiv 
xa( eaxiv and xfjc TB. Sid xd aOxd Sf] xal xo OZ xexpdycovov 
eaxiv xai eaxiv dico xfjc; 6H, xouxeaxiv [dico] xfjc; Ar- xd 
dpa 0Z, Kr xexpdycova duo xcov Ar, TB eiaiv. xal eicel 
laov eaxl xo AH xcp HE, xai eaxi xo AH xo Otto xcov Ar, 
TB- for] yap t\ Hr xfj TB- xal xo HE dpa laov eaxl iu 
Otto Ar, TB - xd dpa AH, HE iaa eaxl xcp 81c; Otto xcov 
AF, TB. eaxi Be xal xd OZ, TK xexpdycova djco xcov AF, 
TB- xd dpa xeaaapa xd 9Z, TK, AH, HE iaa eaxl xolc xe 
duo xcov Ar, TB xexpaycovoic; xal xco Sic Otio xcov Ar, TB 
Ttepie)(ofievcp opiJoycovicp. dXXd xd 8Z, TK, AH, HE oXov 
eaxl xo AAEB, o eaxiv dico xfjc; AB xexpdycovov xo dpa 
duo xfjc; AB xexpdycovov laov eaxl xolc; xe and xcov Ar, 
TB xexpaycovoic; xal xcp Sic utto xcov Ar, TB Ttepie)(o(!evcp 
opiJoycovicp. 

Edv dpa euiJeTa ypa^f] x^irj'dfj, cbc; exu^ev, xo aTto xfjc 




For let the straight-line AB have been cut, at random, 
at (point) C. I say that the square on AB is equal to 
the (sum of the) squares on AC and CB, and twice the 
rectangle contained by AC and CB. 

For let the square ADEB have been described on AB 
[Prop. 1.46], and let BD have been joined, and let CF 
have been drawn through C, parallel to either of AD or 
EB [Prop. 1.31], and let HK have been drawn through 
G, parallel to either of AB or DE [Prop. 1.31]. And since 
CF is parallel to AD, and BD has fallen across them, the 
external angle CGB is equal to the internal and opposite 
(angle) ADB [Prop. 1.29]. But, ADB is equal to ABD, 
since the side BA is also equal to AD [Prop. 1.5]. Thus, 
angle CGB is also equal to GBC. So the side BC is 
equal to the side CG [Prop. 1.6]. But, CB is equal to 
GK, and CG to KB [Prop. 1.34]. Thus, GK is also equal 
to KB. Thus, CGKB is equilateral. So I say that (it is) 
also right-angled. For since CG is parallel to BK [and the 
straight-line CB has fallen across them], the angles KBC 
and GCB are thus equal to two right-angles [Prop. 1.29]. 
But KBC (is) a right-angle. Thus, BCG (is) also a right- 
angle. So the opposite (angles) CGK and GKB are also 
right-angles [Prop. 1.34]. Thus, CGKB is right-angled. 
And it was also shown (to be) equilateral. Thus, it is a 
square. And it is on CB. So, for the same (reasons), 
HF is also a square. And it is on HG, that is to say [on] 
AC [Prop. 1.34]. Thus, the squares HF and KC are 
on AC and CB (respectively). And the (rectangle) AG 
is equal to the (rectangle) GE [Prop. 1.43]. And AG is 
the (rectangle contained) by AC and CB. For GC (is) 
equal to CB. Thus, GE is also equal to the (rectangle 
contained) by AC and CB. Thus, the (rectangles) AG 
and GE are equal to twice the (rectangle contained) by 
AC and CB. And HF and GK are the squares on AC 
and CB (respectively). Thus, the four (figures) HF, GK, 
AG, and GE are equal to the (sum of the) squares on 



53 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



okr\z xexpdytovov iaov eaxl xou; xe duo x65v x[iT]|i.dxGJv xe- 
xpaywvoic; xal xfi 81? utio twv x^ir)[idx«v Tiepiexo^evw op-do- 
yoviw- oTiep e5ei 8eTc;ai. 



^4C and BC, and twice the rectangle contained by AC 
and CB. But, the (figures) HF, CK, AG, and GE are 
(equivalent to) the whole of ADEB, which is the square 
on AB. Thus, the square on AB is equal to the (sum 
of the) squares on AC and CB, and twice the rectangle 
contained by AC and CB. 

Thus, if a straight-line is cut at random then the 
square on the whole (straight-line) is equal to the (sum 
of the) squares on the pieces (of the straight-line), and 
twice the rectangle contained by the pieces. (Which is) 
the very thing it was required to show 



t This proposition is a geometric version of the algebraic identity: (a + b) 2 = a 2 + b 2 + 2 a b. 



'Edv eOifteTa ypa^r] "W^fi £ ^ ^ Ga xal dviaa, to utio twv 
dviatov xfjc; okr\c x^i7)^dx«v Tiepie)(o^.evov opiDoyoviov jiexa 
xoO aTio xfjc ^sia^u xfiv xo[i«v xexpaywvou iaov eaxl xw 
dTio xfjc f]^iaeiac; xexpayd>v<p. 




EH Z 

Eu'deTa yap Tl< ^ ^ AB xex^fja'dto sic, fiev Taa xaxd xo 
T, sic, 8e aviaa xaxd xo A- Xeyw, oxi xo utio xfiv AA, AB 
Tiepie)(6[ievov 6p$oya>viov (jiexa xou duo xfjc; TA xexpaywvou 
iaov eaxl xo duo xfjc; TB xexpaycovco. 

Avayeypdtfdw yap aTio xfjc; TB xexpdywvov xo TEZB, 
xal £KeC£U)fdo f) BE, xal 8ia ^tev xoO A oTioxepa iwv TE, 
BZ TiapdXXrjXoc; f] AH, 5id Se xoO 9 OTioxepa xSv 

AB, EZ TiapdXXr)Xo<; TidXiv ^/-dco f) KM, xal TtdXiv 8ia xoO A 
OTioxepa xov TA, BM TiapdXXr]Xoc; fix^w f] AK. xal CTieliaov 
eaxl xo T0 TiapaTiXf|pco^a xo 0Z napaTiXrjpw^axi, xoivov 
Tipoaxeia'dw xo AM- oXov dpa xo TM oXw iu AZ 'iaov 
eaxiv. dXXd xo TM iu AA Iaov eaxiv, ctieI xal f] Ar xfj 
TB eaxiv Xor\ % xal xo AA dpa xw AZ iaov eaxiv. xoivov 
Ttpoaxeia'dw xo T0- oXov dpa xo A0 xo MNS^ yvtojxovi 
Iaov eaxiv. dXXd xo A6 xo utio xwv AA, AB eaxiv Xar] 
yap f] A0 xfj AB- xal 6 MNS dpa yvwjiwv I'aoc; eaxl x£> 
utio AA, AB. xoivov itpoaxeiai9M xo AH, o eaxiv iaov xG 
drco xfjc; TA- 6 dpa MNS yv«[i«v xal xo AH I'aa eaxl iu 
utio xGv A A, AB uepiexo^tevw op-fJoycovup xal xw drio xfjc; 



Proposition 5 J 



If a straight-line is cut into equal and unequal (pieces) 
then the rectangle contained by the unequal pieces of the 
whole (straight-line), plus the square on the (difference) 
between the (equal and unequal) pieces, is equal to the 
square on half (of the straight-line). 

A CD B 




EG F 

For let any straight-line AB have been cut — equally at 
C, and unequally at D. I say that the rectangle contained 
by AD and DB, plus the square on CD, is equal to the 
square on CB. 

For let the square CEFB have been described on CB 
[Prop. 1.46], and let BE have been joined, and let DG 
have been drawn through D, parallel to either of CE or 
BE [Prop. 1.31], and again let KM have been drawn 
through H, parallel to either of AB or EF [Prop. 1.31], 
and again let AK have been drawn through A, parallel to 
either of CL or BM [Prop. 1.31]. And since the comple- 
ment CH is equal to the complement HF [Prop. 1.43], 
let the (square) DM have been added to both. Thus, 
the whole (rectangle) CM is equal to the whole (rect- 
angle) DF. But, (rectangle) CM is equal to (rectangle) 
AL, since AC is also equal to CB [Prop. 1.36]. Thus, 
(rectangle) AL is also equal to (rectangle) DF. Let (rect- 
angle) CH have been added to both. Thus, the whole 
(rectangle) AH is equal to the gnomon NOP. But, AH 



54 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



TA xexpaycovo. dXXd 6 MNS yva>(iov xal to AH oXov eaxl 
to TEZB xexpdywvov, 6 eaxiv duo xrjc TB- to dpa bub tuv 
AA, AB Tiepiexo^iEvov op^oycoviov jxexa xou diio xrjc TA 
xexpaycivou laov eaxl xw arco xrjc; TB xeTpaywvw. 

'Edv dpa eO'deux ypa^r] xji.Tj'dfj zlc, laa xal aviaa, xo Otco 
xCSv dviaov xrjc; SXrjc; xur^dxtov Ttepie)(6^evov 6pi9oyMviov 
^.exd xoO anb xrjc; (iexac;u xwv xo^iwv xexpaytovou laov eaxl 
xo duo xrjc; f)(Jioe(ac; xexpayc5va>. onep eBei BeT^ai. 



is the (rectangle contained) by AD and For DH 

(is) equal to D.B. Thus, the gnomon NOP is also equal 
to the (rectangle contained) by AD and DB. Let LG, 
which is equal to the (square) on CD, have been added to 
both. Thus, the gnomon NOP and the (square) LG are 
equal to the rectangle contained by AD and DB, and the 
square on CD. But, the gnomon NOP and the (square) 
LG is (equivalent to) the whole square CEFB, which is 
on CB. Thus, the rectangle contained by AD and DB, 
plus the square on CD, is equal to the square on CB. 

Thus, if a straight-line is cut into equal and unequal 
(pieces) then the rectangle contained by the unequal 
pieces of the whole (straight-line), plus the square on the 
(difference) between the (equal and unequal) pieces, is 
equal to the square on half (of the straight-line). (Which 
is) the very thing it was required to show. 



t Note the (presumably mistaken) double use of the label M in the Greek text. 

t This proposition is a geometric version of the algebraic identity: ab + [(a + 6)/2 - b] 2 = [(a + b)/2] 2 . 



f '. 

'Edv eui9eTa ypa^r] "W^fi <^X a ; TCpoaxeiSfj 8s xu; auxfj 
eu'deTa in eCWteiat;, xo bnb xrjc; oXrjc; auv xfj Ttpoaxei^tevr] xal 
xrjc 7tpoaxei[iivr)<; Tiepiexojievov op-fJoytoviov ^texd xoO duo 
xrjc; f\\u.Gsiac, xexpaycjvou '(gov eaxl x£> duo xrjc; auyxei(ievr)<; 
ex xe xrjc; fpiaeiac; xal xrjc; 7ipoaxei^evr)<; xexpaywvcp. 



A r B A 




E HZ 

EO'deTa yap xic f] AB xexjirja'dw 8(/a xaxd xo T orp-siov, 
Tipoaxeia'dw Be xic auxfj euiSela in eCWteiac f] BA- Xeyw, 
oxi xo utto xaiv AA, AB uepie/ojievov opiSoywviov [icxd 
xou anb xrjc; TB xexpaywvou I'aov eaxl xfi and xrjc; IA xe- 
xpaywvw. 

Avayeypdcp'dw yap anb xrjc; TA xexpdywvov xo TEZA, 
xal eTiec^eux'Sw rj AE, xal Bid jiev xou B arjjieiou onoxepa 
xSv Er, AZ TtapdXXiqXoc fj/iSw f] BH, Bid Be xou O arjjieiou 
onoxepa x£>v AB, EZ TTapdXXrjXoc fj/iSw f) KM, xal exi Bid 
xou A onoxepa xfiv TA, AM TiapdXXrjXoc fi/iSw f] AK. 

'End ouv Tar) eaxlv f) Ar xfj TB, I'aov eaxl xal xo AA 



Proposition 6+ 

If a straight-line is cut in half, and any straight-line 
added to it straight-on, then the rectangle contained by 
the whole (straight-line) with the (straight-line) having 
being added, and the (straight-line) having being added, 
plus the square on half (of the original straight-line), is 
equal to the square on the sum of half (of the original 
straight-line) and the (straight-line) having been added. 



A C B ] 


D 
M 






.0/ 

/ \ 
/ i 


K L 


N / 
/ P 


/ 

/ 

/ 



E G F 



For let any straight-line AB have been cut in half at 
point C, and let any straight-line BD have been added to 
it straight-on. I say that the rectangle contained by AD 
and DB, plus the square on CB, is equal to the square 
on CD. 

For let the square CEFD have been described on 
CD [Prop. 1.46], and let DE have been joined, and 
let BG have been drawn through point B, parallel to 
either of EC or DF [Prop. 1.31], and let KM have 
been drawn through point H, parallel to either of AB 
or EF [Prop. 1.31], and finally let AK have been drawn 



55 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



tu rO. dXXd to TO to ©Z faov eaxiv. xal to AA dpa iu 
6Z eaxiv faov. xoivov Tipoaxeiado to TM- oXov apa to 
AM to NSO yvo^tovi eaxiv faov. dXXd to AM soti to utio 
tov AA, AB- far) yap ecrav f] AM xfj AB- xal 6 NSO dpa 
yvo^tov faoc taxi xo utio xov A A, AB [Tiepiex°^ v( P opiJo- 
yovio]. xoivov Tipoaxdtxdo to AH, 6 saxiv faov xo aTio 
xfjc Br xsxpayovo- to dpa utio xov AA, AB Tiepie)(6jj£vov 
opi&oyoviov [lstol tou dno xfjc TB xsxpayovou faov laxl 
xo NSO yvo^ovi xal ifi AH. dXXd 6 NSO yvojiov xal 
to AH oXov eraxl to TEZA xexpdyovov, o eaxt,v aTio xfjc 
TA- to dpa utio xov AA, AB nepi.exo\ievov op-doyoviov 
^texd tou arco xfjt; TB xexpayovou faov eaxl to and xfjc FA 
xexpayovo. 

'Edv dpa eu'dsla ypa^if] Tji.rj'dfj Btya, Tipoaxe'dfj 8s tic 
auxfj eu'dsla etc' eu'deiac, to utio Tfjt; oXr)c auv xfj Ttpo- 
ox£\.\±evr] xal xfjc Ttpoaxa^iivric Tiepiexo^iEvov op-doyoviov 
^.exd xou arco xfjt; fjfiiaeiac xexpayovou faov eaxl to duo 
Tfjt; auyxei^evr)c ex xe xfjc f)fiiae[ac xal Tfjc Tipoaxei^evr]c 
xexpayovo- ojcep e5ei Sel^ai. 



through A, parallel to either of CL or DM [Prop. 1.31]. 

Therefore, since AC is equal to CB, (rectangle) AL is 
also equal to (rectangle) CH [Prop. 1.36]. But, (rectan- 
gle) CH is equal to (rectangle) HF [Prop. 1.43]. Thus, 
(rectangle) AL is also equal to (rectangle) HF. Let (rect- 
angle) CM have been added to both. Thus, the whole 
(rectangle) AM is equal to the gnomon NOP. But, AM 
is the (rectangle contained) by AD and DB. For DM is 
equal to DB. Thus, gnomon NOP is also equal to the 
[rectangle contained] by AD and DB. Let LG, which 
is equal to the square on BC, have been added to both. 
Thus, the rectangle contained by AD and DB, plus the 
square on CB, is equal to the gnomon NOP and the 
(square) LG. But the gnomon NOP and the (square) 
LG is (equivalent to) the whole square CEFD, which is 
on CD. Thus, the rectangle contained by AD and DB, 
plus the square on CB, is equal to the square on CD. 

Thus, if a straight-line is cut in half, and any straight- 
line added to it straight-on, then the rectangle contained 
by the whole (straight-line) with the (straight-line) hav- 
ing being added, and the (straight-line) having being 
added, plus the square on half (of the original straight- 
line), is equal to the square on the sum of half (of the 
original straight-line) and the (straight-line) having been 
added. (Which is) the very thing it was required to show. 



t This proposition is a geometric version of the algebraic identity: (2 a + b) b + a 2 = {a + b) 2 . 



'Edv eu'dsla ypa^tuf] Turj-dfj, oc exu)(ev, to drco Tfjc oXrjc 
xal to dtp' evoc xov xur^dxov xd auvajicpoxepa xexpdyova 
faa ecm to xe 81c utio Tfjc oXrjc xal tou eipr^evou xuf]^axoc 
Tiepiexo[ievo opifayovio xal to arco tou XoitioO xuf^axoc 
xexpayovo. 

a r b 



/ 

/ 

1 

1 


A / 

N / 

/\ 
/ \ 


K / 
/ M 


H / 



A N E 

EuiDeTa yap tic rj AB xexjirfa'do, oc exu)(ev, xaxd to F 
orj^eibv Xeyo, oxi Ta duo xov AB, Br xexpdyova faa eaxl 
to xe 8lc utio tov AB, Br Tiepiexo^tEvo opifayovio xal to 



Proposition 7* 

If a straight-line is cut at random then the sum of 
the squares on the whole (straight-line), and one of the 
pieces (of the straight-line), is equal to twice the rectan- 
gle contained by the whole, and the said piece, and the 
square on the remaining piece. 



H 



/ 

I 

1 


L / 


K / 
/ M 


G 



D N 

For let any straight-line AB have been cut, at random, 
at point C. I say that the (sum of the) squares on AB and 
BC is equal to twice the rectangle contained by AB and 



5G 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



&7i6 Tfjc; TA TETpayGvo). 

Avay£ypdcpiL>Gj yap a7l:o T ^ AB TCTpdyiovov to AAEB- 
xal xaTayeypdcp^co TO a X^ a - 

'End ouv Taov eaxl to AH tG HE, xolvov TtpoaxeicrdM 
to TZ- oXov apa to AZ oXco tG TE Taov eotiv Ta apa 
AZ, TE SmXaaia soti tou AZ. dXXa Ta AZ, TE 6 KAM 
taxi yvGjicov xal to rZ TETpdycovov 6 KAM apa yvGjiGJv 
xal to rZ SiTtXdaid eoti tou AZ. sgti 8e tou AZ SiTtXdaiov 
xal to 81? utto tGv AB, Br- Tar) yap f) BZ Tfj Br- 6 apa 
KAM yvGjjiiov xal to TZ TSTpdytovov iaov eoti tG 81c; utio 
tGv AB, Br. xoivov Tipoaxeio'dw to AH, 6 sgtiv duo Tfjc; 
Ar TCTpdyiovov 6 apa KAM yvG^tov tal Ta BH, HA 
TSTpdywva Taa sotl tG ts 61c; utio tGv AB, Br Ttepi£)(ojj£v« 
op^oycovlcp xal tG arco t^c; Ar TETpayGvip. dXXa 6 KAM 
yvco^cov xal Ta BH, HA TSTpdytova oXov sotI to AAEB xal 
to rZ, a eoTiv duo tGv AB, Br TSTpdywva- Ta apa duo xwv 
AB, Br TETpdywva laa sotI tG [ts] 51c; utio tGv AB, Br 
TterpiEX ^" 6p{>oywviw ^tSTa tou and Tfjc Ar TSTpayGvou. 

'Edv apa euifkTa ypa^jif] T[ir] , df), Gc; stu)(£v, to duo 
Tfjc; 6Xr)<; xal to dtp' evoc; tGv T^ir^dTtov Ta auva^tcpoTspa 
TETpdywva Xaa scttI iu ts 81c; utio Tfjc; oXf]z xal tou 
eipm&vou T[ifj^i.aToc; Ttepie)(o^iv« op'doywvlw xal tG duo 
tou XoitioO TjifjpcToc; TSTpayGvcp- ouep eSei Belial.. 



BC, and the square on CA. 

For let the square ADEB have been described on AB 
[Prop. 1.46], and let the (rest of) the figure have been 
drawn. 

Therefore, since (rectangle) AG is equal to (rectan- 
gle) GE [Prop. 1.43], let the (square) CF have been 
added to both. Thus, the whole (rectangle) AF is equal 
to the whole (rectangle) CE. Thus, (rectangle) AF plus 
(rectangle) CE is double (rectangle) AF. But, (rectan- 
gle) AF plus (rectangle) CE is the gnomon KLM, and 
the square CF. Thus, the gnomon KLM, and the square 
CF, is double the (rectangle) AF. But double the (rect- 
angle) AF is also twice the (rectangle contained) by AB 
and BC. For BF (is) equal to BC. Thus, the gnomon 
KLM, and the square CF, are equal to twice the (rect- 
angle contained) by AB and BC. Let DC, which is 
the square on AC, have been added to both. Thus, the 
gnomon KLM, and the squares BG and GD, are equal 
to twice the rectangle contained by AB and BC, and the 
square on AC. But, the gnomon KLM and the squares 
BG and GD is (equivalent to) the whole of ADEB and 
CF, which are the squares on AB and BC (respectively) . 
Thus, the (sum of the) squares on AB and BC is equal 
to twice the rectangle contained by AB and BC, and the 
square on AC. 

Thus, if a straight-line is cut at random then the sum 
of the squares on the whole (straight-line), and one of 
the pieces (of the straight-line), is equal to twice the rect- 
angle contained by the whole, and the said piece, and the 
square on the remaining piece. (Which is) the very thing 
it was required to show. 



t This proposition is a geometric version of the algebraic identity: (a + b) 2 + a 2 = 2(a + b)a + b 2 . 



*)'• 

'Edv sO'deia ypa^f] T^trydfj, Gc; stu)(£v, to TSTpdxic; Otto 
Tfjc; bXf]z xal evoc; tGv Tur^aTMv Ttspisxo^evov op-doyGviov 
^jLETa tou duo tou Xomou Tuf}pn:oc; TSTpayGvou Taov sgtI 
tG duo ts Tfjc; bXf]z xal tou eipr}\±evov T^uaToc; Gc; drco 
^iiac; dvaypacpevTi TETpayGvw. 

EO'deTa yap tic; f] AB T£Tjif]ai9M, Gc; stu/sv, xaTa to 
r ar^elov Xeyw, oti to TSTpdxu; Otto tGv AB, BT rce- 
pie)(6^.£vov op'doyGviov ^.STa tou duo Tfjc; AT TSTpayGvou 
iaov so"tI tG duo Tfjc; AB, BT Gc; duo [iiaq dvaypacpevTi 
TSTpayGvw. 

'ExpepXf]OTf)w yap in eG'ddac; [xfj AB eu'dela] f) BA, 
xal xeioi9« Tfj TB lot) f] BA, xal dvayeypdcp-dw djio Tfjc; 
A A TSTpdywvov to AEZA, xal xaTayeypdcp'dw SitcXouv to 



Proposition 8* 

If a straight-line is cut at random then four times the 
rectangle contained by the whole (straight-line), and one 
of the pieces (of the straight-line), plus the square on the 
remaining piece, is equal to the square described on the 
whole and the former piece, as on one (complete straight- 
line) . 

For let any straight-line AB have been cut, at random, 
at point C. I say that four times the rectangle contained 
by AB and BC, plus the square on AC, is equal to the 
square described on AB and BC, as on one (complete 
straight-line) . 

For let BD have been produced in a straight-line 
[with the straight-line AB], and let BD be made equal 
to CB [Prop. 1.3], and let the square AEFD have been 
described on AD [Prop. 1.46], and let the (rest of the) 
figure have been drawn double. 



57 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



A 



r b 



A 



B 



M 











H 




1 

\ 




t<t v 

i 




n 




/ Y 







N 
O 



E e A z 

'End ouv Tar) eaxlv f] TB xfj BA, dXXd f] (jiev TB xfj HK 
eaxiv Tar), f] 8e BA xfj KN, xal f] HK apa xfj KN eaxiv Tar). 
8id xd auxd Sf) xal f) nP xrj PO eaxiv iar). xal end Iar) eaxlv 
f] Br XT] BA, f] 8e HK xfj KN, Taov apa eaxl xal xo [lev 
TK xG KA, xo 8s HP xG PN. dXXd xo TK ifi PN eaxiv 
Taov napan;Xr]pw|jiaxa yap xou TO 7tapaXXr]Xoypd|jijj.ou- xal 
xo KA dpa xG HP Taov eaxiv xd xeaaapa dpa xd AK, TK, 
HP, PN Taa aXXrjXoic; eaxiv. xd xeaaapa apa xexpanXdaid 
eaxi xou TK. TtdXiv inei iar) eaxlv f) TB xfj BA, dXXd f) [lev 
BA xrj BK, xouxeaxi xrj TH I'ar), f) 8e TB xfj HK, xouxeaxi 
xfj Hn, eaxiv iar), xal f) TH apa xfj HEI I'ar) eaxlv. xal excel 
I'ar) eaxlv f] \xsm TH xfj Hn, f] Se nP xfj PO, Taov eaxl xal xo 
^tev AH iw Mn, xo 8e nA xG PZ. dXXd xo Affl xG nA eaxiv 
Taov TtapaTiXrjpG^axa yap xou MA icapaXXr]Xoypd|i^ou- xal 
xo AH apa xG PZ Taov eaxiv xd xeaaapa apa xd AH, Mn, 
nA, PZ Taa dXXf]Xoi<; eaxlv xd xeaaapa apa xou AH eaxi 
xexpanXdaia. e8d)fdr] 8e xal xd xeaaapa xd TK, KA, HP, 
PN xou TK xexpaicXdaia- xd apa oxxG, a icepiexei xov ETT 
yvG^tova, xexpauXdaid eaxi xou AK. xal inei xo AK xo uko 
xGv AB, BA eaxiv Iar) yap f] BK xfj BA- xo apa xexpdxu; 
utco xGv AB, BA xexpaicXdaiov eaxi xou AK. eBeix'dr) Se 
xou AK xexpauXdaioc; xal 6 STT yvGjicov xo apa xexpdxu; 
utco xGv AB, BA Taov eaxl xG STT yvG^tovi. xoivov Tcpo- 
axeurdo xo S0, 6 eaxiv Taov xG aTco xrjc Ar xexpayGvor xo 
dpa xexpdxic; utco xGv AB, BA Tcepie)(6^evov op'doyGviov 
^texd xou duo Ar xexpayGvou laov eaxl xG STT yvG^iovi 
xal xG S0. dXXd 6 STT yvG[i«v xal xo S0 6Xov eaxl xo 
AEZA xexpdywvov, o eaxiv and xrjc AA - xo apa xexpdxi; 
utco xGv AB, BA ^texd xou duo Ar Taov eaxl xG and AA 
xexpayGvcp- Tar] 8e f] BA xfj Br. xo apa xexpdxu; utio xGv 
AB, Br icepie)(6(ievov op'doyGviov [icxd xou aTco Ar xe- 
xpayGvou I'aov eaxl xG and xfj; AA, xouxeaxi xG duo xrjc; 
AB xal Br Gc drco [iidc; dvaypacpevxi xexpayGvw. 

'Edv dpa eu'dda ypaji^tf) x^trydfj, G; exu)(ev, xo xexpdxn; 
utco xfj? okr\z xal evbz xGv x^ir)^dxwv Tiepiexo^ievov opiJoyG- 
viov ^texd xou dico xou Xoitcou x^tf][iaxoc xexpayGvou laou 



M 
O 



/ 

/ 


G 




/ 




K \ 

\ 

i 


s / 
/ u 


Q 


R/' 



D 

N 
P 



E H L 

Therefore, since CB is equal to BD, but CB is equal 
to GK [Prop. 1.34], and BD to KN [Prop. 1.34], GK is 
thus also equal to KN. So, for the same (reasons), QR is 
equal to RP. And since BC is equal to BD, and GK to 
KN, (square) CK is thus also equal to (square) KD, and 
(square) GR to (square) RN [Prop. 1.36]. But, (square) 
CK is equal to (square) RN. For (they are) comple- 
ments in the parallelogram CP [Prop. 1.43]. Thus, 
(square) KD is also equal to (square) GR. Thus, the 
four (squares) DK , CK, GR, and RN are equal to one 
another. Thus, the four (taken together) are quadruple 
(square) CK. Again, since CB is equal to BD, but BD 
(is) equal to BK — that is to say, CG — and CB is equal 
to GK — that is to say, GQ — CG is thus also equal to GQ. 
And since CG is equal to GQ, and QR to RP, (rectan- 
gle) AG is also equal to (rectangle) MQ, and (rectangle) 
QL to (rectangle) RF [Prop. 1.36]. But, (rectangle) MQ 
is equal to (rectangle) QL. For (they are) complements 
in the parallelogram ML [Prop. 1.43]. Thus, (rectangle) 
AG is also equal to (rectangle) RF. Thus, the four (rect- 
angles) AG, MQ, QL, and RF are equal to one another. 
Thus, the four (taken together) are quadruple (rectan- 
gle) AG. And it was also shown that the four (squares) 
CK, KD, GR, and RN (taken together are) quadruple 
(square) CK. Thus, the eight (figures taken together), 
which comprise the gnomon STU, are quadruple (rect- 
angle) AK. And since AK is the (rectangle contained) 
by AB and BD, for BK (is) equal to BD, four times the 
(rectangle contained) by AB and BD is quadruple (rect- 
angle) AK. But the gnomon STU was also shown (to 
be equal to) quadruple (rectangle) AK. Thus, four times 
the (rectangle contained) by AB and BD is equal to the 
gnomon STU. Let OH, which is equal to the square on 
AC, have been added to both. Thus, four times the rect- 
angle contained by AB and BD, plus the square on AC, 
is equal to the gnomon STU, and the (square) OH. But, 



58 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



eaxl xfi dtTio xe xfjc okt]Q xal tou eipr^evou xjarpaxoc; tic, the gnomon STU and the (square) OH is (equivalent to) 
duo dvaypacpevxi xexpaycovcp- oTiep eSei SeT^ai. the whole square AEFD, which is on AD. Thus, four 

times the (rectangle contained) by AB and BD, plus the 
(square) on AC, is equal to the square on AD. And BD 
(is) equal to BC. Thus, four times the rectangle con- 
tained by AB and BC, plus the square on AC, is equal to 
the (square) on AD, that is to say the square described 
on AB and BC, as on one (complete straight-line). 

Thus, if a straight-line is cut at random then four times 
the rectangle contained by the whole (straight-line), and 
one of the pieces (of the straight-line), plus the square 
on the remaining piece, is equal to the square described 
on the whole and the former piece, as on one (complete 
straight-line) . (Which is) the very thing it was required 
to show. 

t This proposition is a geometric version of the algebraic identity: 4 (a + b) a + b 2 = [(a + 6) + a] 2 . 



•Q'. 

'Edv eu'deTa YP°W^ 1 W^Ti ''■ aa xat dviaa, xd dfto 
x£>v dviawv xfj? oXrjc xur^dxtov xexpdycova SiTiXdaid eoxi 
xoO xe duo xfjc; r^iadac xod xou dTto xfjc ^exa^u iwv xo^wv 
xexpaycovou. 



E 




A TAB 

EO'deTa ydp xi? f] AB xex^a'fiw eu; |iev i'aa xaxd xo 
T, sic, Be aviaa xaxd xo A- Xeyto, oxl xd dno xfiv AA, AB 
xexpdyiova SiTiXdaid eaxi xfiv dno xGv AT, TA xexpaycovcov. 

"H/iSio yap duo xou T xfj AB Tipog opiEtdc; f) TE, xal 
xeicrdio iot) exaxepa xwv AT, TB, xal eTte^eux-dioaav ai EA, 
EB, xal Sid ^iev xou A xfj ET TiapdXXr]Xoc; rjx$« f) AZ, Bid 
Be xou Z xrj AB f) ZH, xai £7i£^eu)(i9« f) AZ. xal ensl lay] 
eaxiv f) Ar xrj TE, iar) eaxl xal i) utio EAT ytovia xfj utio 
AET. xal eiiel op^rj eaxiv f) Tipoc; x« T, Xoinal apa ai utio 
EAT, AET [iia. 6pi9fj I'aai eiaiv xai eiaiv iaar r)[iiaeia apa 
6pi9rjc; eaxiv exaxepa x£Sv utio TEA, TAE. 8ia xd auxd 8f) 
xal exaxepa xwv utio TEB, EBT fjjjiiaeid eaxiv 6pi9f)<;- oXr) 
apa f) utio AEB 6p®r) eaxiv. xal euel f) utio HEZ f]uiaeid 
eaxiv 6pi9rj<;, op-Qy] Be f) utio EHZ- lot} ydp eaxi xfj evxoc xal 
aTtevavxiov xfj utio ETB- XoiTtf] apa f) utio EZH f)|iiaeid eaxiv 



Proposition Qt 

If a straight-line is cut into equal and unequal (pieces) 
then the (sum of the) squares on the unequal pieces of the 
whole (straight-line) is double the (sum of the) square 
on half (the straight-line) and (the square) on the (dif- 
ference) between the (equal and unequal) pieces. 



E 




A C D B 

For let any straight-line AB have been cut — equally at 
C, and unequally at D. I say that the (sum of the) squares 
on AD and DB is double the (sum of the squares) on AC 
and CD. 

For let CE have been drawn from (point) C, at right- 
angles to AB [Prop. 1.11], and let it be made equal to 
each of AC and CB [Prop. 1.3], and let EA and EB 
have been joined. And let DF have been drawn through 
(point) D, parallel to EC [Prop. 1.31], and (let) FG 
(have been drawn) through (point) F, (parallel) to AB 
[Prop. 1.31]. And let AF have been joined. And since 
AC is equal to CE, the angle EAC is also equal to the 
(angle) AEC [Prop. 1.5]. And since the (angle) at C is 
a right-angle, the (sum of the) remaining angles (of tri- 
angle AEC), EAC and AEC, is thus equal to one right- 



59 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



op'dfjc;- lor] dpa [eaxlv] f] uno HEZ ycovia xfj uno EZH- mots 
xal nXeupa f] EH xfj HZ eaxiv '(or). ndXiv etcei f) npog xS B 
ywvia f)^iaeid eaxiv opi&fjc;, 6p-&y) 8e f) uno ZAB- lot) yap 
ndXiv eaxl xfj evxoc; xal dnevavxiov xfj uno ErB- Xomf] apa 
f| utio BZA f]^uaeid eaxiv op'dfjc;- iar) apa f] npoc; iS B yovia 
xrj uno AZB- waxe xal nXeupa f] ZA nXeupa xrj AB eaxiv 
tar), xal enel iar] eaxlv f) Ar xfj TE, iaov eaxl xal xo dno Ar 
iw duo TE- xa apa duo x£>v Ar, TE xexpdytova BmXdaid 
eaxi xou dno Ar. xou; 8e dno xwv Ar, TE laov eaxl xo dno 
xfjc; EA xexpdytovov 6pi9f) yap f] Ono ArE ywvia- xo apa 
duo xfjc; EA 5mXdaiov eaxi xou duo xfjc; Ar. ndXiv, enel Iar] 
eaxlv f) EH xfj HZ, 'laov xal xo duo xfj? EH xG duo xfjc; HZ- 
xa apa duo xwv EH, HZ xexpdywva BmXdaid eaxi xou dno 
xfjc; HZ xexpaytovou. xou; 8e duo xwv EH, HZ xexpayovou; 
'laov eaxl xo duo xfjc EZ xexpdywvov xo apa dno xfjc EZ 
xexpdywvov BmXdaiov eaxi xou dno xfj? HZ. I'ar) 8e f] HZ 
xr] TA- xo apa duo xfjc EZ SmXdaiov eaxi xou dno xfjc TA. 
eaxi Be xal xo dno xfj? EA SmXdaiov xou dno xfjc; Ar- xa 
apa duo xwv AE, EZ xexpdywva BmXdaid eaxi xwv dno xSv 
Ar, TA xexpaywvwv. xolc Be dno x£>v AE, EZ I'aov eaxl 
xo dno xfjc; AZ xexpdyovov op-dfj yap eaxiv f) uno AEZ 
ywvia- xo apa dno xfjc; AZ xexpdywvov 8mXdaiov eaxi xwv 
dno xc5v Ar, TA. xo 8e dno xfjc; AZ 1'aa xd dno x«v AA, 
AZ- op-df) yap f] npoc xo A ywvia- xa apa dno xov AA, AZ 
BmXdaid eaxi xwv dno xtov Ar, TA xexpaywvwv. Tar) 8e f) 
AZ xfj AB- xa apa dno xwv AA, AB xexpdywva BmXdaid 
eaxi iwv dno xwv Ar, TA xexpdywvwv. 

'Eav apa eu'dela ypa^jif] x^irydfj eic i'aa xal dviaa, xa dno 
x£>v dviawv xfjc; oXrjc; x[ir]|jidx«v xexpdywva BmXdaid eaxi 
xou xe dno xfjc; r^iaeiac xal xou dno xfjc; piexacu xov xo^ifiv 
xexpaycovou- onep eBei SeT^ai. 



angle [Prop. 1.32]. And they are equal. Thus, (angles) 
CEA and CAE are each half a right-angle. So, for the 
same (reasons), (angles) CEB and EEC are also each 
half a right-angle. Thus, the whole (angle) AEB is a 
right-angle. And since GEF is half a right-angle, and 
EGF (is) a right-angle — for it is equal to the internal and 
opposite (angle) ECB [Prop. 1.29] — the remaining (an- 
gle) EFG is thus half a right-angle [Prop. 1.32]. Thus, 
angle GEF [is] equal to EFG. So the side EG is also 
equal to the (side) GF [Prop. 1.6]. Again, since the an- 
gle at B is half a right-angle, and (angle) FDB (is) a 
right-angle — for again it is equal to the internal and op- 
posite (angle) ECB [Prop. 1.29] — the remaining (angle) 
BFD is half a right-angle [Prop. 1.32]. Thus, the angle at 
B (is) equal to DFB. So the side FD is also equal to the 
side DB [Prop. 1.6]. And since AC is equal to CE, the 
(square) on AC (is) also equal to the (square) on CE. 
Thus, the (sum of the) squares on AC and CE is dou- 
ble the (square) on AC. And the square on EA is equal 
to the (sum of the) squares on AC and CE. For angle 
ACE (is) a right-angle [Prop. 1.47]. Thus, the (square) 
on EA is double the (square) on AC. Again, since EG 
is equal to GF, the (square) on EG (is) also equal to 
the (square) on GF. Thus, the (sum of the squares) on 
EG and GF is double the square on GF. And the square 
on EF is equal to the (sum of the) squares on EG and 
GF [Prop. 1.47]. Thus, the square on EF is double the 
(square) on GF. And GF (is) equal to CD [Prop. 1.34]. 
Thus, the (square) on EF is double the (square) on CD . 
And the (square) on EA is also double the (square) on 
AC. Thus, the (sum of the) squares on AE and EF is 
double the (sum of the) squares on AC and CD. And 
the square on AF is equal to the (sum of the squares) 
on AE and EF. For the angle AEF is a right-angle 
[Prop. 1.47]. Thus, the square on AF is double the (sum 
of the squares) on AC and CD. And the (sum of the 
squares) on AD and DF (is) equal to the (square) on 
AF. For the angle at D is a right-angle [Prop. 1.47]. 
Thus, the (sum of the squares) on AD and DF is double 
the (sum of the) squares on AC and CD. And DF (is) 
equal to DB. Thus, the (sum of the) squares on AD and 
DB is double the (sum of the) squares on AC and CD. 

Thus, if a straight-line is cut into equal and unequal 
(pieces) then the (sum of the) squares on the unequal 
pieces of the whole (straight-line) is double the (sum of 
the) square on half (the straight-line) and (the square) on 
the (difference) between the (equal and unequal) pieces. 
(Which is) the very thing it was required to show. 



t This proposition is a geometric version of the algebraic identity: a 2 + b 2 = 2[([a + b]/2) 2 + ([a + b]/2 - b) 2 ]. 



60 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



i . 

'Edv eu'de'ia ypa^jif] x[j.r]'df) Bixa, Kpoaxe-dfj 8s xi; auxfj 
eu-dela etc' eu-deia;, to duo xfj; oX/]; auv xfj Kpoaxei^ievr] 
xal xo dico xfj; Kpoaxeiuevr); xd auva^icpoxepa xexpdywva 
SijcXdaid eaxi xou xe dico xfj; f^iaeia; xal xou dico xfj; auy- 
xei[ievr); ex xe xfj; f]^iaeia; xal xfj; Kpoaxeiuevr]; 6c; and 
[lid; dvaypacpevxo; xexpaycovou. 



E Z 




H 

Eu'de'ia yap xi; f) AB xex^.f|a , dtL> 8[)(a xaxd xo T, Ttpo- 
axeia^co Se xi; auxfj eu'de'ia etc' eMeia; f) BA- Xeyw, oxi xd 
dico xwv AA, AB xexpdyiova SmXaaid eaxi xSv dico iSv 
AT, TA xexpaywvwv. 

'"H/iJcd yap duo xou T ar)|jieiou xfj AB Kpo; op-da; i) TE, 
xal xeicdoy Xar\ exaxepa xwv Ar, TB, xal eite^eu)cd«aav 
ai EA, EB- xal 5id ^iev xou E xfj AA KapdXXrjXo; fjx$w r) 
EZ, 8ia 8e xou A xfj TE KapdXXrjXo; fjx'&M i) ZA. xal etceI 
sic; KapaXXf]Xou; eu'deia; xd; Er, ZA eu'deia xi; eveKeaev 
f] EZ, ai uico TEZ, EZA dpa §ualv op-ddi; I'aai eiaiv ai 
apa uko ZEB, EZA 8uo op-dGv eXdaaove; eiaiv ai 8e 
die' eXaaaovwv fj Suo opiDwv expaXXo^tevai au^iKiKxouaiv 
ai apa EB, ZA expaXXo^ievai era xd B, A ^ep/] au\i- 
iceaouvxai. expepXf|aT!)waav xal aujiKiKxexwaav xaxd xo H, 
xal eTreCeu)fdw r] AH. xal eicel Tar) eaxlv f) Ar xfj TE, Xar] 
eaxi xal ywvia r] uko EAr xfj uko AEr- xal op-Qy) f) Kpo; iu 
T- f]^iaeia apa op'dfjc; [eaxiv] exaxepa xwv utco EAr, AEr. 
Bid xd auxa Sf] xal exaxepa xwv uko TEB, EBr f)^uaeid eaxiv 
op'dfjc;- opiDf] apa eaxlv f\ uko AEB. xal ctccI f)[iiaeia opiDfj; 
eaxiv f) uko EBr, f^iaeia apa op'dfjc; xal f] uko ABH. eaxi 
Se xal f] uko BAH 6pi!)rj- tar) yap eaxi xfj uko ATE- evaXXdc^ 
yap - Xomf] apa f] uko AHB f^iaeid eaxiv op'dfjc;- f) apa uko 
AHB xfj uko ABH eaxiv i'ar]- waxe xal KXeupd f] B A KXeupa 
xfj HA eaxiv Xar\. KaXiv, CKel f) uko EHZ f^iaeid eaxiv 
opiDfjc;, op'df) 8e f) Kpo; xw Z- i'ar] yap eaxi xfj aKevavxiov xfj 
Kpo; xw r- Xomf] apa i) uko ZEH f^iaeid eaxiv op'dfj;- Tar] 
apa f) uko EHZ ywvia xfj uko ZEH- waxe xal KXeupd f] HZ 
KXeupa xfj EZ eaxiv lot}, xal eKel [Tar] eaxlv f) Er xfj TA], 
Taov eaxi [xal] xo aKO xfj; Er xexpdywvov xw aKo xfj; TA 



Proposition 10* 

If a straight-line is cut in half, and any straight-line 
added to it straight-on, then the sum of the square on 
the whole (straight-line) with the (straight-line) having 
been added, and the (square) on the (straight-line) hav- 
ing been added, is double the (sum of the square) on half 
(the straight-line), and the square described on the sum 
of half (the straight-line) and (straight-line) having been 
added, as on one (complete straight-line). 



E F 




For let any straight-line AB have been cut in half at 
(point) C, and let any straight-line BD have been added 
to it straight-on. I say that the (sum of the) squares on 
AD and DB is double the (sum of the) squares on AC 
and CD. 

For let CE have been drawn from point C, at right- 
angles to AB [Prop. 1.11], and let it be made equal to 
each of AC and CB [Prop. 1.3], and let EA and EB have 
been joined. And let EF have been drawn through E, 
parallel to AD [Prop. 1.31], and let FD have been drawn 
through D, parallel to CE [Prop. 1.31]. And since some 
straight-line EF falls across the parallel straight-lines EC 
and FD, the (internal angles) CEF and EFD are thus 
equal to two right-angles [Prop. 1.29]. Thus, FEB and 
EFD are less than two right-angles. And (straight-lines) 
produced from (internal angles whose sum is) less than 
two right-angles meet together [Post. 5]. Thus, being pro- 
duced in the direction of B and D, the (straight-lines) 
EB and FD will meet. Let them have been produced, 
and let them meet together at G, and let AG have been 
joined. And since AC is equal to CE, angle EAC is also 
equal to (angle) AEC [Prop. 1.5]. And the (angle) at 
C (is) a right-angle. Thus, EAC and AEC [are] each 
half a right-angle [Prop. 1.32]. So, for the same (rea- 
sons), CEB and EBC are also each half a right-angle. 
Thus, (angle) AEB is a right-angle. And since EBC 
is half a right-angle, DBG (is) thus also half a right- 
angle [Prop. 1.15]. And BDG is also a right-angle. For 
it is equal to DCE. For (they are) alternate (angles) 



Gl 



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ELEMENTS BOOK 2 



xExpaycivcp- xd apa (xko xo5v Er, TA xsxpdyova SiTtXdaid 
saxi toO duo xfjc TA xsxpaycbvou. xou; 8s anb t«v Er, FA 
Taov sax! xo duo xfjc; EA- xo apa duo xfjc; EA xsxpdywvov 
8iTtXdaiov saxi xou arco xfjc; Ar xsxpayiovou. TtdXiv, she! Xar\ 
sax!v f) ZH xfj EZ, Taov sax! xa! xo djio xfjc ZH x£> duo xfjc 
ZE- xa apa and xwv HZ, ZE SmXdaid saxi xou arco xfj? EZ. 
xolc 8s duo xwv HZ, ZE taov sax! xo duo xfjc; EH' xo apa 
aito xfjc; EH 8i7tXdai6v saxi xou and xfjc; EZ. Tar] 8s f] EZ xfj 
TA- xo apa duo xfjc; EH xsxpdytovov 8i7tXdai6v saxi xou arco 
xfjc; TA. £8e()fdir] Ss xa! xo diio xfjc; EA 8i7tXdaiov xou arco 
xfjc; Ar- xd apa duo xwv AE, EH xsxpdycova 8mXdaid saxi 
iwv diio iwv Ar, TA xsxpaywvwv. xoTc 8s ano xwv AE, 
EH xsxpaywvoic Taov eax! xo duo xfjc; AH xsxpdycovov xo 
apa duo xfjc AH 8i7tXdai6v saxi xwv duo xwv Ar, TA. x£> 
8s and xfjc; AH Taa eax! xa duo xwv AA, AH- xa apa arco 
xwv AA, AH [xexpdywva] 8i7tXdaid saxi xwv anb xtov Ar, 
EA [xexpay u-vov] . Tar] 8e f] AH xfj AB- xd apa aTio iSv 
AA, AB [xsxpdyova] 8i7iXdaid saxi xwv duo x£Sv Ar, TA 
xsxpaywvwv. 

'Edv apa sO'dsTa ypa^f] xjirj'dfj 8i)(a, TipoaxsiDfj 8s xic 
auxfj su'dsTa z% sui!)s(ac, xo arco xfjc; oXr]c auv xfj irpo- 
axsi^isvr] xa! xo duo xfjc; Ttpoaxsi^svr)c xd auva^icpoxspa 
xsxpdywva SiTtXdaid saxi xou xs duo xfjc; f](jiiasiac xa! xou 
d:i:6 xfjc auyxsi^svr]c sx xs xfjc; fjuiasiac xa! xfjc; irpo- 
oxsi\±evr}z cbc duo [iiclc, dvaypacpsvxoc xsxpayovou- oTtsp 
s8si 8sle;ai. 



[Prop. 1.29]. Thus, the remaining (angle) DGB is half 
a right-angle. Thus, DGB is equal to DBG. So side BD 
is also equal to side GD [Prop. 1.6]. Again, since EGF is 
half a right-angle, and the (angle) at F (is) a right-angle, 
for it is equal to the opposite (angle) at C [Prop. 1.34], 
the remaining (angle) FEG is thus half a right-angle. 
Thus, angle EGF (is) equal to FEG. So the side GF 
is also equal to the side EF [Prop. 1.6]. And since [EC 
is equal to CA] the square on EC is [also] equal to the 
square on CA. Thus, the (sum of the) squares on EC 
and CA is double the square on CA. And the (square) 
on EA is equal to the (sum of the squares) on EC and 
CA [Prop. 1.47]. Thus, the square on EA is double the 
square on AC. Again, since FG is equal to EF, the 
(square) on FG is also equal to the (square) on F E. 
Thus, the (sum of the squares) on GF and FE is dou- 
ble the (square) on EF. And the (square) on EG is equal 
to the (sum of the squares) on GF and FE [Prop. 1.47]. 
Thus, the (square) on EG is double the (square) on EF. 
And EF (is) equal to CD [Prop. 1.34]. Thus, the square 
on EG is double the (square) on CD. But it was also 
shown that the (square) on EA (is) double the (square) 
on AC. Thus, the (sum of the) squares on AE and EG is 
double the (sum of the) squares on AC and CD. And the 
square on AG is equal to the (sum of the) squares on AE 
and EG [Prop. 1.47]. Thus, the (square) on AG is double 
the (sum of the squares) on AC and CD. And the (sum 
of the squares) on AD and DC is equal to the (square) 
on AG [Prop. 1.47]. Thus, the (sum of the) [squares] on 
AD and DG is double the (sum of the) [squares] on AC 
and CD. And DG (is) equal to DB. Thus, the (sum of 
the) [squares] on AD and DB is double the (sum of the) 
squares on AC and CD. 

Thus, if a straight-line is cut in half, and any straight- 
line added to it straight-on, then the sum of the square 
on the whole (straight-line) with the (straight-line) hav- 
ing been added, and the (square) on the (straight-line) 
having been added, is double the (sum of the square) on 
half (the straight-line), and the square described on the 
sum of half (the straight-line) and (straight-line) having 
been added, as on one (complete straight-line). (Which 
is) the very thing it was required to show. 



t This proposition is a geometric version of the algebraic identity: (2 a + b) 2 + b 2 = 2 [a 2 + (a + ft) 2 ]. 

ia'. Proposition 11+ 

Tf]v 8o$sTaav su-dslav xs^sTv waxs xo utto xfjc; oXrjc xa! To cut a given straight-line such that the rectangle 
xou sxspou xfiv x^rjjidxMv TtspiS)(6[isvov opiSoytoviov Taov contained by the whole (straight-line), and one of the 
slvai x« duo xou Xoittou x(ifpaxoc; xsxpaywvw. pieces (of the straight-line), is equal to the square on the 

remaining piece. 



G2 



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T K A 

'Eaxco f] 6o r deiaa euiJela f) AB- BsT 8/) xfjv AB xejisiv 
waxe to utio xfjc oXrjc; xal tou exepou xGv x^ir^dxtov 
TC£pi£X ^ evov op'doycoviov Taov e!vai xG cxtio tou XoittoO 
x^ir^axoc; xexpaytovto. 

AvayeypdcpiJo yap omo xfjc; AB xexpdytovov xo ABAr, 
xal xex^f]ai9to f) Ar 8[)(a xaxd xo E ar^eiov, xal STre^sux-dto 
f] BE, xal SiVjx'S" #] TA era xo Z, xal xdcrdco xfj BE Xor\ f) 
EZ, xal dvayeypacpi9co arco xfjc AZ xexpdycovov xo Z6, xal 
8if])(Tf)w f] H0 era xo K- Xeyco, oxi f] AB xex^Tjxai xaxd xo 6, 
&>ax£ xo Otto xfiv AB, B6 Tiepiex o K £vov opftoycoviov Taov 
ttoieiv xtp arco xfjc; AO xexpaytovcp. 

Tkel yap eru'deTa f) Ar xex^trjxai Si/a xaxd xo E, 
rcpoaxeixai 8e auxfj f] ZA, xo dpa Otto xcov TZ, ZA tis- 
piexo^tEvov opiJoyMviov [isxd xou aTio xfjc; AE xexpaywvou 
laov saxl xo arco xfjc EZ xexpaycovco. iar] 8e f] EZ xfj EB- 
xo dpa Otto xov TZ, ZA ^.exd xou and xfjc; AE I'aov eaxl 
xfi duo EB. dXXd xw drco EB I'aa eaxl xd duo x£Sv BA, 
AE- opftf] yap f) Tipoc; ifi A ytovia- xo dpa utto xfiv TZ, 
ZA jisxd xou duo xfjc; AE I'aov saxl xolc; arco xcov BA, AE. 
xoivov dcp/]pr]OTf)w xo duo xfjc; AE- Xoittov dpa xo utto xfiv 
TZ, ZA Tiepiexo^ievov op'doycoviov laov eaxl iw dico xfjc; AB 
xsxpaytovto. xai eaxi xo jisv utto xiov TZ, ZA xo ZK- Iar) 
yap f) AZ xfj ZH- xo 8e arco xfjc; AB xo AA- xo dpa ZK Taov 
eaxl xtp AA. xoivov dpr)pf|a , dco xo AK- Xoittov dpa xo Z6 
xfii 0A laov eaxiv. xa( eraxi xo [lev 0A xo utto xcov AB, 
B0- iar) yap f] AB xfj BA- xo 8e Z6 xo arco xfjc; A9- xo 
dpa utto xfiv AB, B6 Ttspis/o^ievov opiSoytoviov I'aov eaxl 
x£> drco 0A xexpaycovco. 

H dpa 8oi9eTaa eMela f] AB xexpirjxai xaxd xo coaxe 
xo Otto xcov AB, B0 Trepiexo^ievov 6pi9oycoviov laov ttoisTv 
to aTTO xfjc; 0A xexpaywvw- orcep e8ei Tioifjaai. 




C K D 

Let AB be the given straight-line. So it is required to 
cut AB such that the rectangle contained by the whole 
(straight-line), and one of the pieces (of the straight- 
line), is equal to the square on the remaining piece. 

For let the square ABDC have been described on AB 
[Prop. 1.46], and let AC have been cut in half at point 
E [Prop. 1.10], and let BE have been joined. And let 
CA have been drawn through to (point) F, and let EF 
be made equal to BE [Prop. 1.3]. And let the square 
FH have been described on AF [Prop. 1.46], and let GH 
have been drawn through to (point) K. I say that AB has 
been cut at H such as to make the rectangle contained by 
AB and BR equal to the square on AH. 

For since the straight-line AC has been cut in half at 
E, and FA has been added to it, the rectangle contained 
by CF and FA, plus the square on AE, is thus equal to 
the square on EF [Prop. 2.6]. And EF (is) equal to EB. 
Thus, the (rectangle contained) by CF and FA, plus the 
(square) on AE, is equal to the (square) on EB. But, 
the (sum of the squares) on BA and AE is equal to the 
(square) on EB. For the angle at A (is) a right-angle 
[Prop. 1.47]. Thus, the (rectangle contained) by CF and 
FA, plus the (square) on AE, is equal to the (sum of 
the squares) on BA and AE. Let the square on AE have 
been subtracted from both. Thus, the remaining rectan- 
gle contained by CF and FA is equal to the square on 
AB. And FK is the (rectangle contained) by CF and 
FA. For AF (is) equal to FG. And AD (is) the (square) 
on AB. Thus, the (rectangle) FK is equal to the (square) 
AD. Let (rectangle) AK have been subtracted from both. 
Thus, the remaining (square) FH is equal to the (rectan- 
gle) HD. And HD is the (rectangle contained) by AB 
and BH. For AB (is) equal to BD. And FH (is) the 
(square) on AH. Thus, the rectangle contained by AB 



G3 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



and BH is equal to the square on HA. 

Thus, the given straight-line AB has been cut at 
(point) H such as to make the rectangle contained by 
AB and BH equal to the square on HA. (Which is) the 
very thing it was required to do. 

t This manner of cutting a straight-line — so that the ratio of the whole to the larger piece is equal to the ratio of the larger to the smaller piece — is 
sometimes called the "Golden Section". 



IP'. 

'Ev xoT; djj.pXuy«v[oi; xpiywvoi; to duo xf)<; xfjv 
d^pXelav ywviav uuoxeivoua/]; uXeupa; xexpdyiovov jiel^ov 
eaxi t«v duo xGv xf]v d[ipXeTav ywviav uepiexouafiiv 
uXeupCSv xexpayGvojv xfi uepiexo^ievcp Su; Ouo xe [iia; xwv 
uepl xr]v d^pXeTav ycoviav, ecp' f]v f] xdi&exo; uiuxei, xai xfj; 
duoXajipavo^evT]; exxo; (mo xfj; xa-dexou upo; xfj djipXeia 
ycovia. 



B 




A a r 

"Eaxco djjipXuYcdviov xpiywvov xo ABr d^pXelav e/ov 
xr)V Ouo BAr, xai fix^" duo xoO B ar]^efou era xf)v EA 
expX/j'daaav xdiJexo; f] BA. Xeyw, oxi xo duo xfj; Br 
xexpdycovov \is%ov eaxi xwv duo xov BA, Ar xexpaycovov 
xo 51? Ouo xwv EA, AA uepiexo^evcp op-doywviw. 

'Euel yap eu'dela f] EA xex|ir]xai, <i>; exuxev, xaxd xo A 
arjjj.elov, xo dpa duo xfj? Ar 1'aov eaxi xoT; duo xtov EA, 
AA xexpaycovoi; xai xfi 51? Ouo xfiv EA, AA uepiexojievip 
6pi9oya>v[<j. xoivov upoaxeiaiJo xo duo xfj? AB' xd dpa 
duo x£>v TA, AB I'aa eaxl xoT; xe duo xCSv LA, AA, AB xe- 
xpaycovoi; xai xtp 61c; Ouo xCSv BA, AA [uepiexo^ievti> opiSo- 
ycovicp]. dXXd xol; ^tev duo xov TA, AB Taov eaxl xo duo 
xfj; TB- op-df] yap f) upo; xto A ywvia- xoT? 8e duo xwv A A, 
AB i'aov xo duo xfj; AB- xo dpa duo xfj; TB xexpdywvov 
Taov eaxl xoT; xe duo xwv TA, AB xexpaycovoi; xai xco 51; 
Ouo xcov BA, AA uepiexo^ievcp op^oycovicp- oaxe xo duo xfj; 
LB xexpdycovov xcov duo xcov TA, AB xexpaywvov ^icTCov 
eaxi tu 81; Ouo xcov BA, AA uepiexo^evco opiSoycovicp. 

'Ev dpa xoT; d^pXuycovioi; xpiycovoi; xo duo xfj; xf]v 
d^pXelav ycoviav Ouoxeivooar]? uXeopd; xexpdycovov y.s%6\i 
eaxi ifiv duo xcov xf)V d^tpXeTav ycoviav uepiexooacov 



Proposition 12+ 

In obtuse-angled triangles, the square on the side sub- 
tending the obtuse angle is greater than the (sum of the) 
squares on the sides containing the obtuse angle by twice 
the (rectangle) contained by one of the sides around the 
obtuse angle, to which a perpendicular (straight-line) 
falls, and the (straight-line) cut off outside (the triangle) 
by the perpendicular (straight-line) towards the obtuse 
angle. 



B 




DA C 

Let ABC be an obtuse-angled triangle, having the an- 
gle BAC obtuse. And let BD be drawn from point B, 
perpendicular to CA produced [Prop. 1.12]. I say that 
the square on BC is greater than the (sum of the) squares 
on BA and AC, by twice the rectangle contained by CA 
and AD. 

For since the straight-line CD has been cut, at ran- 
dom, at point A, the (square) on DC is thus equal to 
the (sum of the) squares on CA and AD, and twice the 
rectangle contained by CA and AD [Prop. 2.4]. Let the 
(square) on DB have been added to both. Thus, the (sum 
of the squares) on CD and DB is equal to the (sum of 
the) squares on CA, AD, and DB, and twice the [rect- 
angle contained] by CA and AD. But, the (square) on 
CB is equal to the (sum of the squares) on CD and DB. 
For the angle at D (is) a right-angle [Prop. 1.47]. And 
the (square) on AB (is) equal to the (sum of the squares) 
on AD and DB [Prop. 1.47]. Thus, the square on CB 
is equal to the (sum of the) squares on CA and AB, and 
twice the rectangle contained by CA and AD. So the 
square on CB is greater than the (sum of the) squares on 



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ELEMENTS BOOK 2 



TtXeupfiv xexpaytovwv xfi Ttepixo^iivw 81? Otto xe [iia? xwv CA and AB by twice the rectangle contained by CA and 
itepl xf]v djipXeTav ywviav, ecp' fjv #] xdiJexo? tuttxcl, xal xfj? AD. 

dmoXa[ipavo^£VT)<; exxo? bnb xfj? xaiiexou rcpo? xfj d^pXeia Thus, in obtuse-angled triangles, the square on the 
ycoviqr ojiep eSei 8eTc;ai. side subtending the obtuse angle is greater than the (sum 

of the) squares on the sides containing the obtuse an- 
gle by twice the (rectangle) contained by one of the 
sides around the obtuse angle, to which a perpendicu- 
lar (straight-line) falls, and the (straight-line) cut off out- 
side (the triangle) by the perpendicular (straight-line) to- 
wards the obtuse angle. (Which is) the very thing it was 
required to show. 

t This proposition is equivalent to the well-known cosine formula: BC 2 = AB 2 + AC 2 - 2 AB AC cos BAC, since cos BAC = —AD/AB. 



*Ev xoT? 6c;uyGJv[oi? xpiycjvoi? xo duo xfj? xrjv 6c;eTav 
ycoviav unoxeivouar]? jtXeupd? xexpdywvov eXaxxov laxi 
xfiv duo x«v xr)v o^elav yioviav nspis/ouawv TtXeupfiv xe- 
xpayiovtov xai Ttepiexo(jievG3 81? Otto xe [iia? xfiv Ttepl xfjv 
6i;eTav ywviav, ecp' f]v f] xd-dexo? TUTtxei, xdi xrjc; dTtoXajjipa- 
vojievr)? evxo? Otto xfjc; xai9exou Ttpo? xfj o^eitx y«v(a. 



A 




BAT 

'Eaxco 6c;uy«viov xpiywvov xo ABr 6c;eTav e^ov xf]v 
Ttpo? iS B ywviav, xdi /jx^" diio xoO A ar^dou Ira xf]v 
Br xd-dexo? f\ AA- Xeyco, oxi xo duo xfjc Ar xexpdycovov 
eXaxxov eaxi xCSv duo xov TB, BA xexpaywvov xc5 Sic; tmb 
xov TB, BA 7i£piex°^E vt P op-doycoviw. 

Tkel yap cuiMa f] TB xex^trjxai, d>? exuxev, xaxd xo 
A, xd dpa omb xfiv TB, BA xexpdycova laa eaxl xw xe 
81? Otto xfiv TB, BA Ttepiexo^tevw op'doywvta) xdi xfii duo 
xfj? Ar xexpaywvo. xoivov upooxeia'dw xo duo xfj? AA 
xexpdytovov xd dpa duo xwv TB, BA, AA xexpdywva laa 
eaxl tu xe 81? utio twv TB, BA uepiexo^evw opi^oyoviw 
xal xoT? dfto xwv AA, Ar xexpaywvio?. dXXa xoT? [lev arto 
iwv BA, AA I'aov xo duo xfj? AB' op'df) yap f] Ttpo? xw A 
ycovia- xoT? 8e duo xwv AA, Ar I'aov xo duo xfj? Ar- xd 
dpa duo xtov TB, BA laa eaxl x£3 xe arco xfj? Ar xal tu 81? 
utco twv TB, BA- waxe ^tovov xo arto xfj? Ar eXaxxov eaxi 



Proposition 13+ 

In acute-angled triangles, the square on the side sub- 
tending the acute angle is less than the (sum of the) 
squares on the sides containing the acute angle by twice 
the (rectangle) contained by one of the sides around the 
acute angle, to which a perpendicular (straight-line) falls, 
and the (straight-line) cut off inside (the triangle) by the 
perpendicular (straight-line) towards the acute angle. 



A 




B D C 

Let ABC be an acute-angled triangle, having the an- 
gle at (point) B acute. And let AD have been drawn from 
point A, perpendicular to BC [Prop. 1.12]. I say that the 
square on AC is less than the (sum of the) squares on 
CB and BA, by twice the rectangle contained by CB and 
BD. 

For since the straight-line CB has been cut, at ran- 
dom, at (point) D, the (sum of the) squares on CB and 
BD is thus equal to twice the rectangle contained by CB 
and BD, and the square on DC [Prop. 2.7]. Let the 
square on DA have been added to both. Thus, the (sum 
of the) squares on CB, BD, and DA is equal to twice 
the rectangle contained by CB and BD, and the (sum of 
the) squares on AD and DC. But, the (square) on AB 
(is) equal to the (sum of the squares) on BD and DA. 
For the angle at (point) D is a right-angle [Prop. 1.47]. 



G5 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



x£5v duo xcov TB, BA xsxpaytovtov xcp 5k uno xfiv TB, BA 
7tepiexo[ji£V(p op-doyMviw. 

'Ev dpa xok o^uytovbic; xpiyiovoic; xo dmo xrjg xr)v o^slav 
ywviav UTtoxeivoua/jt; TtXeup&c; xexpdywvov eXaxxov eaxi 
xwv duo xcov xrjv o^eiav ytoviav rcspiexouatov nXeupfiv xs- 
xpaycoviov xio Ttspis/o^ievcp 5k 0716 xs [iiac, xfiv Txepl xr]v 
o^slav ycoviav, ecp' fjv f) xd-dexoc; tutcxsi, xdi xrjc aTtoXa^pa- 
vojievrjc; svxoc Otco xfjc xaiJexou Ttpoc; xrj o^eia y«v[qc oTtep 
eBei 5dc;ai. 



And the (square) on AC (is) equal to the (sum of the 
squares) on AD and DC [Prop. 1.47]. Thus, the (sum of 
the squares) on CB and BA is equal to the (square) on 
AC, and twice the (rectangle contained) by CB and BD. 
So the (square) on AC alone is less than the (sum of the) 
squares on CB and BA by twice the rectangle contained 
by CB and BD. 

Thus, in acute-angled triangles, the square on the side 
subtending the acute angle is less than the (sum of the) 
squares on the sides containing the acute angle by twice 
the (rectangle) contained by one of the sides around the 
acute angle, to which a perpendicular (straight-line) falls, 
and the (straight-line) cut off inside (the triangle) by 
the perpendicular (straight-line) towards the acute angle. 
(Which is) the very thing it was required to show. 



t This proposition is equivalent to the well-known cosine formula: AC 2 = AB 2 + BC 2 — 2 AB BC cos ABC, since cos ABC = BD /AB. 




r 



J A 



'Eaxto xo 5oi3ev eu-duypa^ov xo A- 5eT 5rj xw A 
eG'duypd^w 1'aov xexpdywvov auaxrpaaTJai. 

Suveaxdxw ydp xQ A eu'duypd^iw urov TtapaXXrjXo- 
ypa^ov op'doywviov xo BA - d \itv ouv iar\ eaxlv f] BE 
xrj EA, yeyovot; dv sir] xo emxax'dev. auveaxaxai ydp x£> 
A eu'duypd^w i'aov xexpdywvov xo BA- ei 5e ou, ^iia xfiv 
BE, EA jiei^cov eaxiv. eaxo ^dCtov f] BE, xdi expepXria'do 
etc! xo Z, xdi xeiaaJw xrj EA for) t\ EZ, xdi xex^trjo'dw f] BZ 
5[)(a xocxd xo H, xdi xevxpa) xw H, 5iaoxf]^axi 5e evl xwv 
HB, HZ f]^ixuxXiov y£ypdcpif)w xo B9Z, xdi expepXf]a'do f) 
AE era xo 6, xdi ineZs^x^ ^ HQ. 

Tkei ouv eO'deia f\ BZ xex^trjxai etc; y.sv foa xaxd xo H, tic, 
be aviaa xaxd xo E, xo dpa Oreo xfiv BE, EZ 7i£pi£x°HE vov 
opiDoywviov ^.exd xou duo xfj<; EH xexpaycovou i'aov eaxi 
tu aTto xfjc HZ xexpaywvw. for] 5s f] HZ xrj HQ- xo dpa 
U7i6 xwv BE, EZ ^texd xoO aito xrj<; HE i'aov lax! iw dfto 
xfjc; HQ. iS 5e duo xfj<; HQ foa eaxi xd duo xwv QE, EH 



D 

Let A be the given rectilinear figure. So it is required 
to construct a square equal to the rectilinear figure A. 

For let the right-angled parallelogram BD, equal to 
the rectilinear figure A, have been constructed [Prop. 
1.45]. Therefore, if BE is equal to ED then that (which) 
was prescribed has taken place. For the square BD, equal 
to the rectilinear figure A, has been constructed. And if 
not, then one of the (straight-lines) BE or ED is greater 
(than the other). Let BE be greater, and let it have 
been produced to F, and let EF be made equal to ED 
[Prop. 1.3]. And let BF have been cut in half at (point) 
G [Prop. 1.10]. And, with center G, and radius one of 
the (straight-lines) GB or GF, let the semi-circle BHF 
have been drawn. And let DE have been produced to H, 
and let GH have been joined. 

Therefore, since the straight-line BF has been cut — 
equally at G, and unequally at E — the rectangle con- 



66 



STOIXEIQN p\ 



ELEMENTS BOOK 2 



xexpdywva - to apa utto tSv BE, EZ \iexa. xou duo HE laa 
eaxl xoT<; aito xfiv 8E, EH. xoivov dcp/jprjo-dcL) to duo xfj<; HE 
xexpdywvov Xomov apa to (mo xov BE, EZ Tiepiexo^evov 
opiSoycjviov I'aov saxl xai duo xrjg EG xsxpaytovto. dXXa xo 
utto xfiv BE, EZ xo BA eaxiv Xar] yap f) EZ xfj EA- xo 
apa BA 7iapaXXr)X6ypa[i^ov iaov soxl xG duo xfjc; 0E xe- 
xpaytovtp. 1'aov 8s xo BA x£i A Euduypdji^co. xal xo A apa 
£ui36ypa|ji[j.ov i'aov sax! xcp duo xfjc EO dvaypacprjao^evw 
xexpaycivw. 

Tw apa So'devxi eu'duypd^cp x£> A laov xexpdywvov 
auveaxaxai xo duo xfjc E6 dvaypacp^ao^ievov onep eSei 
noifjaai. 



tained by BE and SF, plus the square on EG, is thus 
equal to the square on GF [Prop. 2.5]. And GF (is) equal 
to GH. Thus, the (rectangle contained) by BE and EF, 
plus the (square) on GE, is equal to the (square) on GH. 
And the (sum of the) squares on HE and EG is equal to 
the (square) on GH [Prop. 1.47]. Thus, the (rectangle 
contained) by BE and EF, plus the (square) on GE, is 
equal to the (sum of the squares) on HE and EG. Let 
the square on GE have been taken from both. Thus, the 
remaining rectangle contained by BE and EF is equal to 
the square on EH. But, BD is the (rectangle contained) 
by BE and EF. For EF (is) equal to ED. Thus, the par- 
allelogram BD is equal to the square on HE. And BD 
(is) equal to the rectilinear figure A. Thus, the rectilin- 
ear figure A is also equal to the square (which) can be 
described on EH. 

Thus, a square — (namely), that (which) can be de- 
scribed on EH — has been constructed, equal to the given 
rectilinear figure A. (Which is) the very thing it was re- 
quired to do. 



67 



68 



ELEMENTS BOOK 3 

Fundamentals of Plane Geometry Involving 

Circles 



69 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



"Opoi. 

a', 'laoi xuxXoi Eiaiv, Gv ai 8id|iExpoi i'aai siaiv, fj Gv ai 
ex iwv xsvxpov laai eiaiv. 

P'. Eu-dsTa xuxXou sqxxTCXEcrdai XsYSxai, rjxu; aTCxo^isvr) 
xou xuxXou xal sxPaXXo^isvr] ou xe^ivei xov xuxXov. 

y'. KuxXoi scpdTCxscrdai dXXrjXtov Xcyoviai o'lxivsc; aTixo- 
^.svoi dXXrjXwv ou xsjivouaiv dXXiqXouc;. 

8'. 'Ev xuxX« i'aov ditExsiv duo xou xsvxpou su-dslai 
Xsyovxai, oxav ai duo xoO xsvxpou etc'' auxdc xadsxoi 
dyo^ievai '(aai Gaiv. 

e'. MeTCov 8e aTCExsiv XEYSxai, Icp' fjv f] ^isiCwv xdi9sxo<; 

TUTCXEl. 

9'. T^ifj^a xuxXou lax! xo TCEpisxo^tEvov axrj(ia utco xs 
eu-Qsiac, xal xuxXou TCEpicpspsiac;. 

T^r|uaxo<; 8s ywvia saxiv f] TCspisxo^isvr] utco xs 
su'dsia? xal xuxXou TCEpicpspsiac;. 

T)'. 'Ev x^irj^iaxi 8s yavia saxiv, oxav era xfj? TCspi- 
cpspsiac; xou x^rj^axoe; Xrjcp'dfj xi ar)(ieTov xal octc' auxou era 
xa nspaxa xrj<; su'dsia?, fj saxi pdau; xou x^ir^axoc;, stcl- 
^eux^woiv su-dsla!,, f] TCepiexo^Evr] ywvia utco xGv stci^su- 
X'dsiaGv sutJeiGv. 

"Oxav 8e ai Ttspisxouaai x/]v ywviav su'dslai aito- 
Xa^pdvoal xiva TCEpicpspsiav, etc' exei\>r\q Xsysxai psprjXEvai 
f] Ywvia. 

1'. To^isix; 8s xuxXou saxiv, oxav Ttpo<; xG xsvxpG xou 
xuxXou auaxa-dfj Y^via, xo TCEpisxo^svov axrjjia utco xe iSv 
x/]v Y"v(av TiEpiExouawv su-dsiGv xal xfjc aTCoXa^pavo^svr](; 
utc' auxGv TCspicpspsiac;. 

ia'. "O^ioia x[if]^axa xuxXwv Eaxi xa Ssxo^isva ycoviat; 
Iaa<;, fj ev olz ai Y^viai laai dXXr]Xai<; siaiv. 



a . 

Tou 8oi9svxo<; xuxXou xo xsvxpov supslv. 

'Eax« 6 So-dsli; xuxXoc 6 ABE 5sT 8rj xou ABr xuxXou 
xo xsvxpov supsTv. 

Air|x$w xu; si<; auxov, G<; sxuxsv, su'dsTa f) AB, xal 
x£x^tr]OTf)w 8ixa xaxa xo A ar\\ieiov, xal aito xou A xfj AB 
7ipo<; 6pM<; rjx'dw f] Ar xal Sifix^w etc! xo E, xal xsxuf|a , d« 
f] TE 8ixa xaxa xo Z- Xsy«, oxi xo Z xsvxpov saxi xou ABr 
[xuxXou] . 

Mf] yap, dXX' si 5uvax6v, saxw xo H, xal ETiECsux'dwCTav 
ai HA, HA, HB. xal etcei for) saxlv ?) AA xrj AB, xoivr) 8e f) 
AH, 860 8/] ai AA, AH 860 xdi<; HA, AB i'aai siaiv sxaxspa 
sxaxspa- xal pdau; f) HA pdasi xrj HB iaxiv tar)- ex xsvxpou 
Yap- Y«via apa f) utco AAH Y^via xfj utco HAB iaf] saxiv. 



Definitions 

1. Equal circles are (circles) whose diameters are 
equal, or whose (distances) from the centers (to the cir- 
cumferences) are equal (i.e., whose radii are equal). 

2. A straight-line said to touch a circle is any (straight- 
line) which, meeting the circle and being produced, does 
not cut the circle. 

3. Circles said to touch one another are any (circles) 
which, meeting one another, do not cut one another. 

4. In a circle, straight-lines are said to be equally far 
from the center when the perpendiculars drawn to them 
from the center are equal. 

5. And (that straight-line) is said to be further (from 
the center) on which the greater perpendicular falls 
(from the center). 

6. A segment of a circle is the figure contained by a 
straight-line and a circumference of a circle. 

7. And the angle of a segment is that contained by a 
straight-line and a circumference of a circle. 

8. And the angle in a segment is the angle contained 
by the joined straight-lines, when any point is taken on 
the circumference of a segment, and straight-lines are 
joined from it to the ends of the straight-line which is 
the base of the segment. 

9. And when the straight-lines containing an angle 
cut off some circumference, the angle is said to stand 
upon that (circumference) . 

10. And a sector of a circle is the figure contained by 
the straight-lines surrounding an angle, and the circum- 
ference cut off by them, when the angle is constructed at 
the center of a circle. 

11. Similar segments of circles are those accepting 
equal angles, or in which the angles are equal to one an- 
other. 

Proposition 1 

To find the center of a given circle. 

Let ABC be the given circle. So it is required to find 
the center of circle ABC. 

Let some straight-line AB have been drawn through 
(ABC), at random, and let (AB) have been cut in half at 
point D [Prop. 1.9]. And let DC have been drawn from 
D, at right-angles to AB [Prop. 1.11]. And let (CD) have 
been drawn through to E. And let CE have been cut in 
half at F [Prop. 1.9]. I say that (point) F is the center of 
the [circle] ABC. 

For (if) not then, if possible, let G (be the center of the 
circle), and let GA, CD, and GB have been joined. And 
since AD is equal to DB, and DG (is) common, the two 



70 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



oxav 8e euif)eTa in euiJeTav axaiSeTaa xdg ecpe<;rj<; ytovtag 
laac, dXXrjXaic; ranfj, dpfti) exaxepa xaiv lacov yioviGv eaxiv 
6p0r] dpa eaxlv f) Otto HAB. eaxl Be xal f] utto ZAB opTEby 
Tor) apa f) Otto ZAB xfj Otto HAB, f] [ici^mv xrj eXdxxovr 
onep eaxlv dBuvaxov. oux dpa xo H xevxpov eaxl xou ABr 
xuxXou. by.o'\.Gic, Br] Be^ojiev, oxi ouB' dXXo xi TtXf]v xoO Z. 



r 





H 


z 




\ A 





(straight-lines) AD, DC are equal to the two (straight- 
lines) BD, DG^ respectively. And the base GA is equal 
to the base GB. For (they are both) radii. Thus, angle 
ADG is equal to angle GDB [Prop. 1.8]. And when a 
straight-line stood upon (another) straight-line make ad- 
jacent angles (which are) equal to one another, each of 
the equal angles is a right-angle [Def. 1.10]. Thus, GDB 
is a right-angle. And FDB is also a right-angle. Thus, 
FDB (is) equal to GDB, the greater to the lesser. The 
very thing is impossible. Thus, (point) G is not the center 
of the circle ABC. So, similarly, we can show that neither 
is any other (point) except F. 

c 



F 


G 


\ D 




E 



To Z dpa arj^elov xevxpov eaxl xou ABT [xuxXou]. 



Thus, point F is the center of the [circle] ABC. 



riopiajaa. 

'Ex Br) xouxou cpavepov, oxi edv ev xuxXw cu-deld uc, 
eMeldv xiva 8()(a xal upoc; op-ddc; xejivr), Era xfjc; xejivouarjc; 
eaxl xo xevxpov xou xuxXou. — ouep eBei icoirjaai. 



t The Greek text has "GD, DB", which is obviously a mistake. 

P'- 

'Edv xuxXou era xfj? uepicpepeiac; Xrjcp'dfi Buo xu)(6vxa 
a/jueTa, f] era xd ar^eTa era^euyvu^evr) eO'dela evxoc; rceaelxai 
xou xuxXou. 

Tiaxw xuxXo<; 6 ABT, xal era xfj? uepicpepeiac; auxou 
eiXrjcpiEko Buo xu)(6vxa a/jjiela xd A, B- Xeyto, oxi f] duo 
xou A era xo B era^euYvu^evrj euiSela evxog jieaelxai xou 
xuxXou. 

Mr] y^Pj «XX' el 6uvax6v, raTtxexco exxoc <i>c; fj AEB, xal 
eiXricp'dw xo xevxpov xou ABT xuxXou, xal eaxw xo A, xal 
ejceCeux'dwaav al AA, AB, xal 8ir|X$« f] AZE. 

'Etcei ouv lar\ eaxlv f] A A xrj AB, iar\ apa xal ytovta f) 
utco AAE xrj uko ABE- xal excel xpiywvou xou AAE ^iia 



Corollary 

So, from this, (it is) manifest that if any straight-line 
in a circle cuts any (other) straight-line in half, and at 
right-angles, then the center of the circle is on the for- 
mer (straight-line) . — (Which is) the very thing it was 
required to do. 



Proposition 2 

If two points are taken at random on the circumfer- 
ence of a circle then the straight-line joining the points 
will fall inside the circle. 

Let ABC be a circle, and let two points A and B have 
been taken at random on its circumference. I say that the 
straight-line joining A to B will fall inside the circle. 

For (if) not then, if possible, let it fall outside (the 
circle), like AEB (in the figure). And let the center of 
the circle ABC have been found [Prop. 3.1], and let it be 
(at point) D. And let DA and DB have been joined, and 
let DFE have been drawn through. 

Therefore, since DA is equal to DB, the angle DAE 



71 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



TiXeupd TtpoaexpepXrjxai f] AEB, [isi^cov apa f\ hub AEB 
y«v(a xfjc utco AAE. Tar) 8e f] utio AAE xrj (mo ABE- 
^tei^wv apa f] tmo AEB xrj<; O716 ABE. utio 8e xf|V ^tel^ova 
ywviav f) jieii^wv TtXeupd unoxeivei' [iei^iov apa f] AB xfjc; 
AE. Tar] 8e f] AB xrj AZ. jie^wv apa r) AZ xrjc; AE f) 
eXdxxcov xrjc; ^iei^ovoc;- ouep eaxlv d8uvaxov. oux apa f] 
dtTio xou A era xo B eiuCeuyvu^evr] eu'dela exxoc; Tteaelxai 
xoO xuxXou. o^iolwc; 8r] Se^o^isv, oxi ouBe en' auxrjc; xrjc; 
Ttepicpepelac;- evxoc; apa. 




'Eav apa xuxXou em xfjc; nepicpepeiac; X/]cpif)fj 860 xu)(6vxa 
a/)[ieTa, f) era xa ar)[ieTa eTti^euyvu^evr) euiJeTa evxoc; Tteaelxai 
xoO xuxXou- oitep e8ei SeT^ai. 

Y • 

'Eav ev xuxXcp eu^sTa uc, 81a xou xevxpou euiMdv xiva 
\ir\ 8id xoO xevxpou 8i)(a xejivr], xal Ttpoc; op'dac; auxrjv xejivei- 
xal eav Tipog opMc; auxr]v xejivr), xal 8(xa aux/jv xejivei. 

"Eaxa> xuxXoc; 6 ABr, xal ev auxcp eO'deTd xi<; 61a xou 
xevxpou f) TA eMeldv xiva \ly] Bid xou xevxpou xrjv AB Bi/a 
xejivexco xaxd xo Z arj^ielov Xeyco, 6x1 xal Ttpoc; 6pi)ac; auxrjv 
xe^ivei. 

EiXrjCpiftco yap xo xevxpov xou ABr xuxXou, xal eaxco 
xo E, xal £K£^eu)(i9«oav al EA, EB. 

Kod CTtelTar) eaxlv f] AZ xrj ZB, xolvt) Be rj ZE, 860 8ualv 
I'oai [etatv] 1 xal pdaic; f) EA pdaei xrj EB tor) 1 yiovia apa f) 
Oito AZE yiovia xrj Otto BZE Tar) eaxiv. oxav Be eu-dela Itc' 
eu-delav axaOeTaa xdc; ecpec^fjc; ycoviac; Taac dXXfjXaic; Ttoifj, 
opiDr] exaxepa xwv Tacov ywvifiv eaxiv exaxepa apa xwv 
UTto AZE, BZE opftfi eaxiv. i] EA apa Sid xou xevxpou 
ouaa xrjv AB [Li] Bid xou xevxpou ouaav Bixa xepivouaa xal 
Ttpoc; 6pM<; xejivei. 



(is) thus also equal to DBE [Prop. 1.5]. And since in tri- 
angle DAE the one side, AEB, has been produced, an- 
gle DEB (is) thus greater than DAE [Prop. 1.16]. And 
DAE (is) equal to DBE [Prop. 1.5]. Thus, DEB (is) 
greater than DBE. And the greater angle is subtended 
by the greater side [Prop. 1.19]. Thus, DB (is) greater 
than DE. And DB (is) equal to DF. Thus, DF (is) 
greater than DE, the lesser than the greater. The very 
thing is impossible. Thus, the straight-line joining A to 
B will not fall outside the circle. So, similarly, we can 
show that neither (will it fall) on the circumference itself. 
Thus, (it will fall) inside (the circle) . 




Thus, if two points are taken at random on the cir- 
cumference of a circle then the straight-line joining the 
points will fall inside the circle. (Which is) the very thing 
it was required to show. 

Proposition 3 

In a circle, if any straight-line through the center cuts 
in half any straight-line not through the center then it 
also cuts it at right-angles. And (conversely) if it cuts it 
at right-angles then it also cuts it in half. 

Let ABC be a circle, and, within it, let some straight- 
line through the center, CD, cut in half some straight-line 
not through the center, AB, at the point F. I say that 
(CD) also cuts (AB) at right-angles. 

For let the center of the circle ABC have been found 
[Prop. 3.1], and let it be (at point) E, and let EA and 
EB have been joined. 

And since AF is equal to FB, and FE (is) common, 
two (sides of triangle AFE) [are] equal to two (sides of 
triangle BFE). And the base EA (is) equal to the base 
EB. Thus, angle AFE is equal to angle BFE [Prop. 1.8]. 
And when a straight-line stood upon (another) straight- 
line makes adjacent angles (which are) equal to one an- 
other, each of the equal angles is a right-angle [Def. 1.10]. 
Thus, AFE and BFE are each right-angles. Thus, the 



72 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



r 




A 

AXXd 8f] f] TA xrjv AB icpoc; op'dag xejivexav Xeyw, oxi 
xal 8ixa auxrjv xe^tvei, xouxeaxiv, oxi I'ar) eaxlv f] AZ xfj ZB. 

Tt5v yap auxwv xaxaaxeuaai)evxMv, etcei iar\ Eaxlv f) 
EA xfj EB, iar] eaxl xal ywvia f] utxo EAZ xfj uito EBZ. 
eaxl Se xal op-dr) rj Otco AZE op'dfj xfj (mo BZE iar)- 860 
dpa xpiywvd eaxi EAZ, EZB xa<; 860 ywv(a<; 8uol ywviau; 
Iaa<; E^ovxa xal jiiav jcXeupdv [iia TcXeupa I'arjv xoivrjv auxfiv 
xrjv EZ uicoxeivouaav uko ^uav xGv lawv ywviwv xal xa<; 
XoiKag apa TcXeupac; xau; Xomau; icXeupau; Xaac, e^ei- larj apa 
rj AZ xfj ZB. 

'Eav apa ev xuxXw EU'dsTd uc, 81a xou xevxpou euiMdv 
xiva [ufj 81a xou xsvxpou Sixa xe^vrj, xal jcpo<; op^a? auxrjv 
xejuver xal eav Ttpo<; opMc; auxrjv xeuvrj, xal 8[^a auxrjv 
xe^tvei- oicep eSei BeT^ai. 



5'. 

Eav ev xuxXcp 860 eu-deTai xe^ivcoaiv dXXrjXag [urj 8la xou 
xevxpou ouaai, ou xs^vouaiv dXXrjXac; 8[)(a. 

'Eaxco xuxXo<; 6 ABrA, xal ev auxcp 860 eO'delai ai Ar, 
BA xe^ivexMoav dXXrjXa<; xaxa xo E [if] Sid xou xevxpou 
ouaai- Xeyw, oxi ou xejivouaiv dXXf]Xa<; 8[)(a. 

EE yap Suvaxov, xe^vexoaav dXXrjXac; 8[)(a &axe Tarjv 
eTvai xrjv |iev AE xrj Er, xrjv Be BE xrj EA- xal eiXrjcpiL>Gj xo 
xevxpov xou ABrA xuxXou, xal eaxco xo Z, xal ejie^eu)cd« 
f] ZE. 

'EticI ouv eO'deld xig 81a xou xevxpou f] ZE eui&eTdv xiva 
(if) Bid xou xevxpou xrjv Ar 8i/a xejivei, xal itpoc; op-ddc; 
auxrjv xe^ivei- 6p-&y] apa eaxlv fj utto ZEA- ndXiv, creel eu-deld 
tic, fj ZE eu-deldv xiva xrjv BA 8i/a xejivei, xal icpoc; op'da.z 
auxrjv xejivei- op^rj apa rj utto ZEB. e8e£)(i9r) 8e xal fj utto 
ZEA op'drj- iarj apa f) utto ZEA xfj utto ZEB fj eXdxxwv xrj 



(straight-line) CD, which is through the center and cuts 
in half the (straight-line) AB, which is not through the 
center, also cuts (AB) at right-angles. 



c 




D 

And so let CD cut AB at right-angles. I say that it 
also cuts (AB) in half. That is to say, that AF is equal to 

FB. 

For, with the same construction, since EA is equal 
to EB, angle EAF is also equal to EBF [Prop. 1.5]. 
And the right-angle AFE is also equal to the right-angle 
BFE. Thus, EAF and EFB are two triangles having 
two angles equal to two angles, and one side equal to 
one side — (namely), their common (side) EF, subtend- 
ing one of the equal angles. Thus, they will also have the 
remaining sides equal to the (corresponding) remaining 
sides [Prop. 1.26]. Thus, AF (is) equal to FB. 

Thus, in a circle, if any straight-line through the cen- 
ter cuts in half any straight-line not through the center 
then it also cuts it at right-angles. And (conversely) if it 
cuts it at right-angles then it also cuts it in half. (Which 
is) the very thing it was required to show. 

Proposition 4 

In a circle, if two straight-lines, which are not through 
the center, cut one another then they do not cut one an- 
other in half. 

Let ABCD be a circle, and within it, let two straight- 
lines, AC and BD, which are not through the center, cut 
one another at (point) E. I say that they do not cut one 
another in half. 

For, if possible, let them cut one another in half, such 
that AE is equal to EC, and BE to ED. And let the 
center of the circle ABCD have been found [Prop. 3.1], 
and let it be (at point) F, and let FE have been joined. 

Therefore, since some straight-line through the center, 
FE, cuts in half some straight-line not through the cen- 
ter, AC, it also cuts it at right-angles [Prop. 3.3]. Thus, 
FEA is a right-angle. Again, since some straight-line FE 



73 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



jjtei^ovi- oTtep eaxlv d§uvaxov. oux apa ai Ar, BA xe^ivouaiv 




'Eav apa ev xuxXcp Suo eu-delai xe^ivwaiv dXXf|Xa<; ^f] 8la 
xou xevxpou ouaai, ou xe^ivouaiv dXXr|Xa<; Si/a - onep eSei 
BeT^ai. 

S . 

'Eav Buo xuxXoi xe^ivcoaiv dXXr|Xouc, oux eaxai auxcov 
xo auxo xevxpov. 




Ado yap xuxXoi ol ABr, TAH xejivexcoaav dXXr|Xou<; 
xaxd xa B, T ar^ela. Xeyw, on oux eaxai auxGv xo auxo 
xevxpov. 

Ei ydp 8uvax6v, eaxw xo E, xal C7te£eu)(i!}G3 f\ EF, xal 
Bir^-dco f) EZH, ox exu/ev. xal eixei xo E a/jjielov xevxpov 
eaxl xou ABr xuxXou, Tar) eaxlv rj Er xfj EZ. rcaXiv, end xo 
E o/]jj.£lov xevxpov eaxl xou TAH xuxXou, I'ar) eaxlv f] Er 
xfj EH- eBeix^T) Be f] Er xal xfj EZ icny xal f) EZ apa xfj EH 
eaxiv Tar) f] eXdaaov xfj [leiCovi- onep eaxlv dBuvaxov. oux 
apa xo E ar)^eTov xevxpov eaxl xwv ABr, TAH xuxXcov. 

'Eav apa Buo xuxXoi xeuvtoaiv dXXr]Xou<;, oux eaxiv 



cuts in half some straight-line BD, it also cuts it at right- 
angles [Prop. 3.3]. Thus, FEB (is) a right-angle. But 
FEA was also shown (to be) a right-angle. Thus, FEA 
(is) equal to FEB, the lesser to the greater. The very- 
thing is impossible. Thus, AC and BD do not cut one 
another in half. 




Thus, in a circle, if two straight-lines, which are not 
through the center, cut one another then they do not cut 
one another in half. (Which is) the very thing it was re- 
quired to show. 

Proposition 5 

If two circles cut one another then they will not have 
the same center. 




For let the two circles ABC and CDC cut one another 
at points B and C. I say that they will not have the same 
center. 

For, if possible, let E be (the common center), and 
let EC have been joined, and let EFG have been drawn 
through (the two circles), at random. And since point 
E is the center of the circle ABC, EC is equal to EF. 
Again, since point E is the center of the circle CDG, EC 
is equal to EG. But EC was also shown (to be) equal 
to EF. Thus, EF is also equal to EG, the lesser to the 
greater. The very thing is impossible. Thus, point E is not 



74 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



auxwv to auTo xsvxpov onep eSsi SeT^ai. 




Auo yap xuxXoi ol ABr, TAE ecpanxso^woav dXXr]X«v 
xaxa to T CTTjjjieTov Xeyw, oti oux eaxai auTfiv to ai)TO 
xevTpov. 

El yap SuvaTov, eoto to Z, xal ETieCeu/'dw f] Zr, xal 
SirDcdw, £TU)(ev, f] ZEB. 

Tkel ouv to Z ar)[ieTov xevTpov eaxl tou ABr xuxXou, 
Xar] eaTiv r) Zr T/j ZB. TtdXw, ind to Z ar^ieTov xevTpov 
eoxl tou TAE xuxXou, for) saw f) Zr ttj ZE. £5d)cdr] 5e f] 
Zr T/j ZB Xor\- xal f) ZE dpa Tfj ZB eotiv Xai], f) eXAttcov 
Tfj jie^ovi- oTiep saw d5uvaTov. oux dpa to Z ar^eiov 
xevTpov sgtI twv ABr, TAE xuxXiov. 

'Edv dpa 8uo xuxXoi ecpdnTCOVTai dXXr]X«v, oux ecrcai 
auTCSv to auTo xevTpov orcep sSei BsT^ai. 



'Edv xuxXou era xfjt; 8ia^.£Tpou Xrjcp'df) ti ot][ieXo\i, 8 (ir| 
eoTi xevTpov tou xuxXou, arco 8e tou ar\\ie'iov izpbc, tov 
xuxXov TipocnuTTTwaiv su-delal Tiveg, jjlcyicttt) ^iev caTai, ecp' 
rjc; to xevTpov, eXa^iaT/) 8e f) Xomrj, tSv Be dXXov dtel f] 
£YY tov tffc §i a TO ° xevTpou Trj<; duwTspov piei^wv ecrav, 
6uo 5e ^tovov Taai dno tou o/jjisiou TipoaneaouvTai 7tp6<; 
tov xuxXov £cp' exaTspa Trjg EXaxloTrjc;. 



the (common) center of the circles ABC and CDC 

Thus, if two circles cut one another then they will not 
have the same center. (Which is) the very thing it was 
required to show. 

Proposition 6 

If two circles touch one another then they will not 
have the same center. 




For let the two circles ABC and CDE touch one an- 
other at point C. I say that they will not have the same 
center. 

For, if possible, let F be (the common center), and 
let FC have been joined, and let FEB have been drawn 
through (the two circles), at random. 

Therefore, since point F is the center of the circle 
ABC, FC is equal to FB. Again, since point F is the 
center of the circle CDE, FC is equal to FE. But FC 
was shown (to be) equal to FB. Thus, FE is also equal 
to FB, the lesser to the greater. The very thing is impos- 
sible. Thus, point F is not the (common) center of the 
circles ABC and CDE. 

Thus, if two circles touch one another then they will 
not have the same center. (Which is) the very thing it was 
required to show. 

Proposition 7 

If some point, which is not the center of the circle, 
is taken on the diameter of a circle, and some straight- 
lines radiate from the point towards the (circumference 
of the) circle, then the greatest (straight-line) will be that 
on which the center (lies), and the least the remainder 
(of the same diameter) . And for the others, a (straight- 
line) nearer* to the (straight-line) through the center is 
always greater than a (straight-line) further away. And 
only two equal (straight-lines) will radiate from the point 
towards the (circumference of the) circle, (one) on each 



75 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



(side) of the least (straight-line). 

c 



bX \\ / 


z 


E V 1 





'Eaxco xuxXoc 6 ABrA, 8id(iexpo<; Be auxou eaxw f\ A A, 
xal em xfjc A A eiXf|cp'dco ti a/jjieTov to Z, o \ir\ eaxi xevxpov 
xoO xuxXou, xevxpov 8e xou xuxXou eaxw xo E, xal duo xoO 
Z npo? xov ABrA xuxXov TtpoaTimxexcoaav eu'delai xivec od 
ZB, Zr, ZH- Xeyw, oxi ^teyiaxr) \ie\> eaxiv f\ ZA, eXa^iaxr) 
8e f] ZA, xfiv 8e dXXwv f] ^tev ZB xfjc Zr ^tei^cov, f] 8e Zr 
xfjc ZH. 

, EKeCeu)fdwaav yap ai BE, TE, HE. xal eiiel Ttavxoc 
xpiywvou ai 8uo TtXeupal xfjc XoiTtfjc ^teiCovec eiaiv, ad dpa 
EB, EZ xfjc BZ jieiCovec eiaiv. Xor\ 8e f] AE xfj BE [ai dpa 
BE, EZ laai eiai xfj AZ]- ^lel^cov dpa f) AZ xfjc BZ. TtdXiv, 
enel Tar) eaxiv f) BE xfj TE, xoivf) 8e f] ZE, 8uo 6f] al BE, 
EZ Suoi xdic TE, EZ I'oai eiaiv. dXXd xal ycovia f\ uno BEZ 
yoviac xfjc Otto FEZ [ie[£«v pdaic dpa f] BZ pdaeioc xfjc 
rZ ^ei^tov eaxiv. 8id xd auxd 8rj xal f) TZ xfj? ZH ^et^wv 
eaxiv. 

ndXiv, enel ai HZ, ZE xfjc; EH [ici^ovec eiaiv, i'ar] 8e f] 
EH xfj EA, ai dpa HZ, ZE xfjc EA [iei^ovec eiaiv. xoivf) 
dcpflpriadtL) f] EZ- Xomf) dpa f) HZ Xomfjc xfjc; ZA [lei^cov 
eaxiv. (leyiax/) [lev dpa f) ZA, eXa)(iaxr) Se f) ZA, jiei^wv 6e 
f] txev ZB xfjc Zr, f) Be Zr xfjc ZH. 

Aeyio, oxi xal and xou Z arjjieiou 80o ^xovov laai Tipo- 
aiteaouvxai Tipoc xov ABrA xuxXov ecp' exdxepa xfjc ZA 
eXa/iaxrjg. auveaxdxw yap Ttpoc xrj EZ euiJeia xal xai Ttpoc 
auxfj a/j^ieitp xfi E xfj Otto HEZ yojvia I'ar) f) imb ZE9, xal 
eTie^eu)cdw r) Z6. enel ouv la/) eaxiv rj HE xfj E6, xoivf) 
8e f) EZ, 8uo 8f) ai HE, EZ 8uai xdic 6E, EZ laai eiaiv 
xal ywvia f) Otto HEZ ywvia xfj Otto 0EZ lary pdaic dpa 
f] ZH pdaei xfj ZO Tar) eaxiv. Xeyo 5rj, oxi xfj ZH dXXr) 
for] ou KpoaneaeTxai TTpoc xov xuxXov duo xou Z ar^eiou. 
ei yap 8uvax6v, TrpoaTiiTixexo f] ZK. xal euel f\ ZK xfj ZH 
for] eaxiv, dXXd f] ZO xfj ZH [Tar] eaxiv], xal f] ZK dpa xfj 
ZO eaxiv I'ar], f] eyyiov xfjc 8id xou xevxpou xfj ducoxepov 
for)- oTiep dBuvaxov. oux dpa duo xou Z ar)[ieiou exepa xic 




Let ABCD be a circle, and let AZ) be its diameter, and 
let some point F, which is not the center of the circle, 
have been taken on AD. Let E be the center of the circle. 
And let some straight-lines, FB, FC, and FG, radiate 
from F towards (the circumference of) circle ABCD. I 
say that FA is the greatest (straight-line), FD the least, 
and of the others, FB (is) greater than FC, and FC than 
FG. 

For let BE, CE, and GE have been joined. And since 
for every triangle (any) two sides are greater than the 
remaining (side) [Prop. 1.20], EB and EF is thus greater 
than BF. And AE (is) equal to BE [thus, BE and EF 
is equal to AF]. Thus, AF (is) greater than BF. Again, 
since BE is equal to CE, and FE (is) common, the two 
(straight-lines) BE, EF are equal to the two (straight- 
lines) CE, EF (respectively). But, angle BEF (is) also 
greater than angle CEF} Thus, the base BF is greater 
than the base CF. Thus, the base BF is greater than the 
base CF [Prop. 1.24]. So, for the same (reasons), CF is 
also greater than FG. 

Again, since CF and FE are greater than EG 
[Prop. 1.20], and EG (is) equal to ED, GF and FE 
are thus greater than ED. Let EF have been taken from 
both. Thus, the remainder GF is greater than the re- 
mainder FD. Thus, FA (is) the greatest (straight-line), 
FD the least, and FB (is) greater than FC, and FC than 
FG. 

I also say that from point F only two equal (straight- 
lines) will radiate towards (the circumference of) circle 
ABCD, (one) on each (side) of the least (straight-line) 
FD. For let the (angle) FEH, equal to angle GEF, have 
been constructed on the straight-line EF, at the point E 
on it [Prop. 1.23], and let FH have been joined. There- 
fore, since GE is equal to EH, and EF (is) common, 



76 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



Tipoaueaaxai Tipoc; xov xuxXov iar) xfj HZ- (iia apa \i6vr\. 

'Edv apa xuxXou era xfjc; Siauixpou X/jcp'dfj xi arjiMov, 
o (if] eaxi xevxpov xoO xuxXou, drab Be xou ar)[ieiou Ttpoc; 
xov xuxXov TtpoaraTixoaiv cu-dela! xivec;, (xeyiaxrj (xev eaxai, 
ecp' fjc; xo xevxpov, eXaxiaxr] 8s f) XoiTtf], xwv Se dXXwv del f) 
eyyiov xfjc; Sla xou xevxpou xfjc djtwxepov (iei£«v eaxiv, Suo 
Se (iovov I'aai duo xou auxou arpeiou TtpoaTteaoOvxai Ttpoc; 
xov xuxXov ecp' exdxepa xfjc eXa/iaxiqc;- ojtep eSei 8eTi;ai. 



the two (straight-lines) GE, EF are equal to the two 
(straight-lines) HE, EF (respectively). And angle GEF 
(is) equal to angle HEF. Thus, the base FG is equal to 
the base FH [Prop. 1.4]. So I say that another (straight- 
line) equal to FG will not radiate towards (the circumfer- 
ence of) the circle from point F. For, if possible, let FK 
(so) radiate. And since FK is equal to FG, but FH [is 
equal] to FG, FK is thus also equal to FH, the nearer 
to the (straight-line) through the center equal to the fur- 
ther away. The very thing (is) impossible. Thus, another 
(straight-line) equal to GF will not radiate from the point 
F towards (the circumference of) the circle. Thus, (there 
is) only one (such straight-line). 

Thus, if some point, which is not the center of the 
circle, is taken on the diameter of a circle, and some 
straight-lines radiate from the point towards the (circum- 
ference of the) circle, then the greatest (straight-line) 
will be that on which the center (lies), and the least 
the remainder (of the same diameter) . And for the oth- 
ers, a (straight-line) nearer to the (straight-line) through 
the center is always greater than a (straight-line) further 
away. And only two equal (straight-lines) will radiate 
from the same point towards the (circumference of the) 
circle, (one) on each (side) of the least (straight-line). 
(Which is) the very thing it was required to show. 



t Presumably, in an angular sense. 

t This is not proved, except by reference to the figure. 



'Edv xuxXou X/]cpiL>fj xi o/)[i.£lov exxoc;, duo Se xou 
a/][ie(ou 7tp6<; xov xuxXov Sia)(iL>Gaiv eu-delai xivec;, £>v [iia 
[lev 8id xou xevxpou, ai Se XoiTtai, &>c, exu)(ev, xGv (lev upog 
xf]v xoiX/jv mcpicpepeiav TtpoaraTixouacov eu'dsifiiv (leyiaxr] 

eaxiv f] Sid xou xevxpou, xwv Se aXXwv del f\ eyyiov xfjc; 
Sid xou xevxpou xfjc; draoxepov (ie(£«v eaxiv, xwv Se Ttpoc; 
xf]v xupxfjv nepicpepeiav upocnunxouafiv eui&eicov eXa^iaxr) 
[jlev eaxiv f) (iexac;u xou xe ar^efou xal xfjc; Sia^texpou, xwv 
Se aXXov del f] eyyiov t ^ eXayJaxrjc; xfjc; duwxepov eaxiv 
eXdxxcov, Suo Se fiovov laai and xou arpeiou TtpoaTteaoOvxai 
Ttpoc; xov xuxXov ecp' exdxepa xfjc; eXayJaxrjc;. 

"Eaxco xuxXoc; 6 ABr, xal xou ABr eiXfjcp'dw xi ay][ieTov 
exxoc; xo A, xal arc' auxou 8ir])fdwoav eu'delai xivec ai AA, 
AE, AZ, Ar, eaxw Se f] AA Sid xou xevxpou. Xeyto, 
oxi xwv [Lev Tipoc; xf]v AEZr xoiXrjv uepicpepeiav Ttpoara- 
Ttxouawv eu'deiwv (icyiaxr) [Lev eaxiv f) Sid xou xevxpou f) 
AA, pieiCwv Se f] [Lev AE xfjc; AZ f] Se AZ xfjc; Ar, xwv 
Se Tipoc xfjv 0AKH xupxfjv uepicpepeiav TtpoaTtiTtxouaCSv 
eu-deiCSv eXa)([axr) [Lev eaxiv r\ AH f) (lexacu xou ar^fjieiou xal 
xfjc; Siajiexpou xfjc AH, del Se f] eyyiov xfjc; AH eXa)(iaxir]C 
eXdxxwv eaxl xfjc; dicwxepov, f] [Lev AK xfjc; AA, f] Se AA 



Proposition 8 

If some point is taken outside a circle, and some 
straight-lines are drawn from the point to the (circum- 
ference of the) circle, one of which (passes) through 
the center, the remainder (being) random, then for the 
straight-lines radiating towards the concave (part of the) 
circumference, the greatest is that (passing) through the 
center. For the others, a (straight-line) nearer^ to the 
(straight-line) through the center is always greater than 
one further away. For the straight-lines radiating towards 
the convex (part of the) circumference, the least is that 
between the point and the diameter. For the others, a 
(straight-line) nearer to the least (straight-line) is always 
less than one further away. And only two equal (straight- 
lines) will radiate from the point towards the (circum- 
ference of the) circle, (one) on each (side) of the least 
(straight-line). 

Let ABC be a circle, and let some point D have been 
taken outside ABC, and from it let some straight-lines, 
DA, DE, DF, and DC, have been drawn through (the 
circle), and let DA be through the center. I say that for 
the straight-lines radiating towards the concave (part of 



77 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



xfj? A9. 



A 




ElXf](p , dcL) yap to xevxpov xou ABr xuxXou xod eaxto to 
M- xai OTeCEUxtfwaav ai ME, MZ, Mr, MK, MA, M6. 

Kal ind Xai) eaxiv f] AM xfj EM, xoivf] Ttpoaxeia'dco f) 
MA- f) apa AA for) eaxi xai? EM, MA. dXX' ai EM, MA 
xfj? EA jieic^ove? daw xal f) AA apa xfj? EA ^eii^iov eaxiv. 
ndtXiv, end iar) eaxiv f] ME xfj MZ, xoivf] 8e f] MA, ai EM, 
MA apa xai? ZM, MA i'aai Eiaiv xai ytovia f] Otco EMA 
ywvia? xfj? utio ZMA ^.ei^cov eaxiv. pdai? dpa f] EA pdaew? 
xfj? ZA ^teii^ov eaxiv o^xoick Bf) 8eic;o^ev, oxi xai f] ZA xfj? 
EA ^iei^ov eaxiv ^eyiaxr] ^ev dpa f) AA, ^leii^ov 8e f) jiev 
AE xfjt; AZ, f] 8e AZ xfj? Ar. 

Kai etcei ai MK, KA xfj? MA jieiCove? eiaiv, iarj 8s r) 
MH xfj MK, Xomf] dpa f] KA Xonxrj? xfjt; HA jieiCwv eaxiv 
&axe f] HA xfj? KA eXdxxov eaxiv xai enei xpiywvou xou 
MAA em. (jiidt; xwv TtXeupGv xfjt; MA 8uo eu-deTai evxo? 
auveaxd , dr]aav ai MK, KA, ai dpa MK, KA xc5v MA, AA 
eXdxxove? eiaiv Iar] Se f] MK xfj MA- Xoircf] dpa f] AK 
Xomfj? xfj? AA eXdxxcov eaxiv. o^ioico? Bf] Sei^o^tev, oxi 
xai f] AA xfj? A8 eXdxxwv eaxiv eXa)(iaxr) [lev dpa f] AH, 
eXdxxwv Be f] [iev AK xfjt; AA f) 8e AA xfj? A9. 

Aey«, 6x1 xai 860 [lovov i'aai duo xou A ar)[ieiou 
npoaKeaouvxai n:p6? xov xuxXov ecp' exdxepa xfj? AH 
eXa)([axr]?- auveaxdxw rcpo? xfj MA eu'deia xai x£> npbc, 
auxfj ar)[ie[w iS M xfj Otto KMA ywvia i'ar) ywvia f) Otto 
AMB, xai CTieCeux'dw f) AB. xai etc el Xar] eaxiv f] MK xfj 
MB, xoivf] Be f] MA, 860 8f) ai KM, MA 860 xai? BM, MA 



the) circumference, AEFC, the greatest is the one (pass- 
ing) through the center, (namely) AD, and (that) DE (is) 
greater than DF, and DF than DC. For the straight-lines 
radiating towards the convex (part of the) circumference, 
HLKG, the least is the one between the point and the di- 
ameter AG, (namely) DG, and a (straight-line) nearer to 
the least (straight-line) DG is always less than one far- 
ther away, (so that) DK (is less) than DL, and DL than 
than DE. 



D 




For let the center of the circle have been found 
[Prop. 3.1], and let it be (at point) M [Prop. 3.1]. And let 
ME, MF, MC, MK, ML, and MH have been joined. 

And since AM is equal to EM, let MD have been 
added to both. Thus, AD is equal to EM and MD. But, 
EM and MD is greater than ED [Prop. 1.20]. Thus, 
AD is also greater than ED. Again, since ME is equal 
to MF, and MD (is) common, the (straight-lines) EM, 
MD are thus equal to FM, MD. And angle EMD is 
greater than angle FMD} Thus, the base ED is greater 
than the base FD [Prop. 1.24]. So, similarly, we can 
show that FD is also greater than CD. Thus, AD (is) the 
greatest (straight-line), and DE (is) greater than DF, 
and DF than DC. 

And since MK and KD is greater than MD [Prop. 
1.20], and MG (is) equal to MK, the remainder KD 
is thus greater than the remainder GD. So GD is less 
than KD. And since in triangle MLD, the two inter- 
nal straight-lines MK and KD were constructed on one 
of the sides, MD, then MK and KD are thus less than 
ML and LD [Prop. 1.21]. And MK (is) equal to ML. 
Thus, the remainder DK is less than the remainder DL. 
So, similarly, we can show that DL is also less than DH. 
Thus, DG (is) the least (straight-line), and DK (is) less 
than DL, and £>£ than DiJ . 

I also say that only two equal (straight-lines) will radi- 



78 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



Taai eialv exaxepa exaxepqr xal ywvia f| uno KMA ycovia 
xfj UTio BMA iar)- pdaic; apa f) AK pdaei xfj AB for) eaxiv. 
Xeyw [8^] j oti xfj AK eu-deia aXXr] I'ar) ou TipoaTteaeTxai 
7ip6<; xov xuxXov octio xou A ar^eiou. ei yap 8uvax6v, npo- 
aKiKxexo xal eaxw f] AN. end ouv f] AK xfj AN eaxiv Tar), 
dXX' f) AK xfj AB eaxiv Tar), xal f) AB apa xfj AN eaxiv 
Tar), f) eyyiov T ^ AH eXa)((axr](; xfj dn;«xepov [eaxiv] Tary 
orcep dSuvaxov e8e[)fdr]. oux apa rcXeiouc; fj Suo Taai 7ip6<; 
xov ABr xuxXov diio xoO A ar)[ieiou ecp' exdxepa xfj? AH 
eXa)((oxr]<; TtpoaTteaouvxai. 

'Eav apa xuxXou Xrjcp'dfj xi ar)(ieTov exxoc, arco 5e xou 
ay][ie'iou npbc, xov xuxXov 8ia)fd«aiv eu'delai xivec;, Sv ^ua 
^.sv 8id xou xevxpou ai 8e Xomai, (be; exu)(£v, xcov ^iev 7ip6<; 
xf]v xo(Xr)v uepicpepeiav Tipoarauxouawv sutJeiGv (jieyiaxr] 
[lev eaxiv f) 8ia xou xevxou, xfiv 8e aXXwv del f] eyyiov xfjc; 
Sid xou xevxpou xfj<; djiwxepov [ie(C«v eaxiv, xwv 8e 7tp6<; 
xf]v xupxf]v uepicpepeiav KpoaTiiTixouawv euiJeiGv eXa)([axr) 
^ev eaxiv f) ^texa^u xou xe ar^eiou xal xfj? 8ia^texpou, iSv 
8e dXXwv del f] eyyiov xfjc; eXaxiaxrjc xfjc; dftoxepov eaxiv 
eXdxxMv, 8uo 8e ^.ovov Taai duo xou arj^ieiou Tipoaueaouvxai 
7ip6<; xov xuxXov ecp' exdxepa xfjc eXaxiax/]<;- orcep e8ei 
Bel^ai. 



ate from point D towards (the circumference of) the cir- 
cle, (one) on each (side) on the least (straight-line), DG. 
Let the angle DMB, equal to angle KMD, have been 
constructed on the straight-line MD, at the point M on 
it [Prop. 1.23], and let DB have been joined. And since 
MK is equal to MB, and MD (is) common, the two 
(straight-lines) KM, MD are equal to the two (straight- 
lines) BM, MD, respectively. And angle KMD (is) 
equal to angle BMD. Thus, the base DK is equal to the 
base DB [Prop. 1.4]. [So] I say that another (straight- 
line) equal to DK will not radiate towards the (circum- 
ference of the) circle from point D. For, if possible, let 
(such a straight-line) radiate, and let it be DN. There- 
fore, since DK is equal to DN, but DK is equal to DB, 
then DB is thus also equal to DN, (so that) a (straight- 
line) nearer to the least (straight-line) DG [is] equal to 
one further away. The very thing was shown (to be) im- 
possible. Thus, not more than two equal (straight-lines) 
will radiate towards (the circumference of) circle ABC 
from point D, (one) on each side of the least (straight- 
line) DG. 

Thus, if some point is taken outside a circle, and some 
straight-lines are drawn from the point to the (circumfer- 
ence of the) circle, one of which (passes) through the cen- 
ter, the remainder (being) random, then for the straight- 
lines radiating towards the concave (part of the) circum- 
ference, the greatest is that (passing) through the center. 
For the others, a (straight-line) nearer to the (straight- 
line) through the center is always greater than one fur- 
ther away. For the straight-lines radiating towards the 
convex (part of the) circumference, the least is that be- 
tween the point and the diameter. For the others, a 
(straight-line) nearer to the least (straight-line) is always 
less than one further away. And only two equal (straight- 
lines) will radiate from the point towards the (circum- 
ference of the) circle, (one) on each (side) of the least 
(straight-line). (Which is) the very thing it was required 
to show. 



t Presumably, in an angular sense. 

| This is not proved, except by reference to the figure. 



ft'. 

'Eav xuxXou Xrjcp'dfj xi ar\\ie1ov evxoc;, olko 8e xou 
arj^eiou 7ipo<; xov xuxXov Ttpoaiuitxwai nkeiouq fj Suo Taai 
eu-delai, xo Xrj^ev arpeTov xevxpov eaxl xou xuxXou. 

'Eaxco xuxXoc; 6 ABr, evxoc 8e auxou ar)[ieTov xo A, xal 
dTio xou A 7ip6<; xov ABr xuxXov upoaraTixexcoaav TtXeiouc; 
f] 8uo Taai eu-delai al AA, AB, AE Xey«, oxi xo A ar)[ieTov 
xevxpov eaxl xou ABr xuxXou. 



Proposition 9 

If some point is taken inside a circle, and more than 
two equal straight-lines radiate from the point towards 
the (circumference of the) circle, then the point taken is 
the center of the circle. 

Let ABC be a circle, and D a point inside it, and let 
more than two equal straight-lines, DA, DB, and DC, ra- 
diate from D towards (the circumference of) circle ABC. 



79 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



A 




O 

TkeCetijcdcooav yap ai AB, Br xal TeTuVjcydtoaav 
Bi^a xaxd xa E, Z ar^eTa, xal ETuCeux'deiaai al EA, ZA 
8if])(Tf)waav era xa H, K, 0, A ar^ela. 

Tkei ouv igt] eaxiv #] AE xrj EB, xoivr) 8e #] EA, 8uo 
8f) al AE, EA 660 xau; BE, EA laai eialv xal pdau; f] AA 
pdasi xfj AB tar) - y«v(a dpa f) utto AEA ycovia xfj Otto BEA 
iar] eaxlv op'dr] dpa exaxepa xwv O716 AEA, BEA yoviwv f) 
HK dpa xrjv AB xe^ivei Sixa xal npbc, op^dc. xal ejxel, eav 
ev xuxXco EuiDeTd xu; euiDeldv xiva 8ixa xe xal 7tp6<; 6piDd<; 
xe^ivr], £7il xfj<; xe[ivo6ar)<; eaxl xo xevxpov xou xuxXou, era 
xfj? HK dpa eaxl xo xevxpov xou xuxXou. Sid xd auxd 8r) xal 
zid xrjc 0A eaxi xo xevxpov xou ABr xuxXou. xal ouSsv 
exepov xoivov exouaiv ai HK, OA eu'de'iai f\ xo A arpeTov 
xo A dpa ar^elov xevxpov eaxl xou ABr xuxXou. 

'Eav dpa xuxXou XricpiDfj xi ar^elov evxoc, duo 8e xou 
ar\\ieiov izpbz xov xuxXov Ttpoarararwai uXeiouc; rj 860 laai 
eu-delai, xo Xr)cp{>ev ar^elov xevxpov eaxl xou xuxXou- oTtep 
e8ei BeT^ai. 



I say that point D is the center of circle ABC. 



B/ 








F __\ 








k — 


A^ 


D 



H 



For let AB and BC have been joined, and (then) 
have been cut in half at points E and F (respectively) 
[Prop. 1.10]. And ED and FD being joined, let them 
have been drawn through to points G, K, H, and L. 

Therefore, since AE is equal to EB, and ED (is) com- 
mon, the two (straight-lines) AE, ED are equal to the 
two (straight-lines) BE, ED (respectively). And the base 
DA (is) equal to the base DB. Thus, angle AED is equal 
to angle BED [Prop. 1.8]. Thus, angles AED and BED 
(are) each right-angles [Def. 1.10]. Thus, GK cuts AB in 
half, and at right-angles. And since, if some straight-line 
in a circle cuts some (other) straight-line in half, and at 
right-angles, then the center of the circle is on the former 
(straight-line) [Prop. 3.1 corr.], the center of the circle is 
thus on GK. So, for the same (reasons), the center of 
circle ABC is also on HL. And the straight-lines GK and 
HL have no common (point) other than point D. Thus, 
point D is the center of circle ABC. 

Thus, if some point is taken inside a circle, and more 
than two equal straight-lines radiate from the point to- 
wards the (circumference of the) circle, then the point 
taken is the center of the circle. (Which is) the very thing 
it was required to show. 



l . 

KuxXoc; xuxXov ou xejivei xaxd nXebva orj^iela fj 860. 

EE yap 8uvax6v, xuxXog 6 ABT xuxXov xov AEZ 
xe^tvexM xaxd nXeiova ar)[ieTa fj 860 xa B, H, Z, O, xal 
eTuCeuyjdeTaai ai B9, BH Sfya xe^ivecrdwaav xaxd xd K, A 
ar)[ieTa- xal aito xwv K, A xau; B9, BH Ttp6<; opMc; dx^elaai 
ai KT, AM 8ir])cdwoav era xd A, E ar)^eTa. 



Proposition 10 

A circle does not cut a(nother) circle at more than two 
points. 

For, if possible, let the circle ABC cut the circle DEF 
at more than two points, B, G, F, and H . And BH and 
BG being joined, let them (then) have been cut in half 
at points K and L (respectively). And KC and LM be- 
ing drawn at right-angles to BH and BG from K and 
L (respectively) [Prop. 1.11], let them (then) have been 
drawn through to points A and E (respectively). 



80 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 




Tkel ouv ev xuxXw xfi ABr eui)eTd tic, f) AT su'dei&v 
xiva xf]v B9 h'v/O- xal Ttpoc; op-ddc; xe^jivei, era xfjc; Ar apa 
eaxl to xevxpov tou ABr xuxXou. TtdXiv, itnel ev xuxXcp xG 
auxG tu ABr £015616; tic; f] NS eu'deldv xiva xf)v BH 8[)(a 
xal Ttpoc; op'dac; xe^vei, era xfjc; NS apa eaxl xo xevxpov 
xoO ABr xuxXou. eBeix'dr) 8e xal era xfjc; Ar, xal xax' 
oi)8ev au^pdXXouaiv ai Ar, NS eu-delai fj xaxd xo O xo 
O dpa at][ieTov xevxpov laxl xou ABr xuxXou. o^toiwc; Sf] 
8eic;ouev, oxi xal xou AEZ xuxXou xevxpov eaxl xo O 8uo 
apa xuxXwv Teuv6vT«v dXXfjXouc; tGv ABr, AEZ xo auxo 
eaxi xevxpov xo O ojtep eaxlv d56vaxov. 

Oux apa xuxXoc; xuxXov xe^ivei xaxd nXetova arj(ieTa f\ 
8uo- onep e5ei Belial. 



N 









L 







\o 



Therefore, since in circle ABC some straight-line 
AC cuts some (other) straight-line BH in half, and at 
right-angles, the center of circle ABC is thus on AC 
[Prop. 3.1 corr.]. Again, since in the same circle ABC 
some straight-line NO cuts some (other straight-line) BG 
in half, and at right-angles, the center of circle ABC is 
thus on NO [Prop. 3.1 corr.]. And it was also shown (to 
be) on AC. And the straight-lines AC and NO meet at 
no other (point) than P. Thus, point P is the center of 
circle ABC. So, similarly, we can show that P is also the 
center of circle DEF. Thus, two circles cutting one an- 
other, ABC and DEF, have the same center P. The very- 
thing is impossible [Prop. 3.5]. 

Thus, a circle does not cut a(nother) circle at more 
than two points. (Which is) the very thing it was required 
to show. 



lot'. 

'Edv 8uo xuxXoi ecpdnxwvxai dXXf]X«v evToc;, xal Xr)tfdfj 
auxwv xd xevxpa, f) era xd xevxpa auxwv eraCeuyvu^ievr) 
eMela xal exPaXXo^tevr] era xrjv ouvacprjv TieoeTxai tGv 
xuxXgjv. 

Auo yap xuxXoi oi ABT, AAE ecpaTixeo-dwaav dXXf|Xwv 
evxo<; xaxd xo A ar^elov, xal elX^cp-dco xou y.kv ABT xuxXou 
xevxpov xo Z, xou 8e AAE xo H- Xeyco, oxi f\ duo xou H era 
xo Z era^euYvu^evr) eu^ela ex[3aXXo^evr) era xo A neaelxai. 

Mr] yap, dXX' et Suvaxov, ranxexw Gc; f] ZH9, xal 
ejie^eu)cd«aav ai AZ, AH. 

'Eracl ouv ai AH, HZ xfjc; ZA, xouxeaxi xfjc; Z9, (leiCovec; 
eioiv, xoivr) dcprpfjadio f] ZH- XoiTtf] apa f] AH Xomfjc; xfjc 
H0 [is^cov eaxiv. for) 8e f] AH xfj HA- xal f] HA apa 
xfjc; H8 ^teiCwv eaxlv f] eXdxxwv xfjc; [leiCovoc;- oracp eaxlv 
dSuvaxov oux apa f) drab xou Z era xo H eraCeuyvu^ievr) 
eu-dela exxoc; ueaelxai- xaxd xo A apa era xfjc; auvacpfjc 
racaelxai. 



Proposition 11 

If two circles touch one another internally, and their 
centers are found, then the straight-line joining their cen- 
ters, being produced, will fall upon the point of union of 
the circles. 

For let two circles, ABC and ADE, touch one another 
internally at point A, and let the center F of circle ABC 
have been found [Prop. 3.1], and (the center) G of (cir- 
cle) ADE [Prop. 3.1]. I say that the straight-line joining 
G to F, being produced, will fall on A. 

For (if) not then, if possible, let it fall like FGH (in 
the figure), and let AF and AG have been joined. 

Therefore, since AG and GF is greater than FA, that 
is to say FH [Prop. 1.20], let FG have been taken from 
both. Thus, the remainder AG is greater than the re- 
mainder GH. And AG (is) equal to GD. Thus, GD is 
also greater than GH, the lesser than the greater. The 
very thing is impossible. Thus, the straight-line joining F 
to G will not fall outside (one circle but inside the other) . 
Thus, it will fall upon the point of union (of the circles) 



81 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 




'Edv dpa 860 xuxXoi ecpditxcovxai dXXf|XGJv evxoc;, [xal 
Xrjcydrj auxfiv xd xevxpa], f] Ira xd xevxpa auxwv era^eu- 
yvupievr] euiiteTa [xal expaXXojievr)] era xf]v auvacpfjv rceaelxai 
twv xuxXtov ouep e8ei BeT^ai. 

LP'. 

'Eav 860 xuxXoi ecpdnxtovxai dXXf|Xt>)v exxoc;, f] era xa 
xevxpa auxov era^euyvujievT] 6id xfjc eixacpfjc; eXeuaexai. 

B 




E 

Auo yap xuxXoi oi ABr, AAE ecpaTtxea-dtoaav dXXf|XtL>v 
exxoc; xaxd xo A arjjielov, xal elXf]cpi9M xoO [Lev ABT 
xevxpov xo Z, xou 8e AAE xo H' Xeyw, oxi f) drab xoO 
Z era xo H era^euyvunevr) eu-dela 8id xfjt; xaxd xo A eixacpfjc; 
eXeuaexai. 

Mf] yap, dXX'' el 8uvax6v, epxecrdw «<; f) ZrAH, xal 
CTieCeux'Swaav ai AZ, AH. 

Tkel ouv xo Z ar)[ieTov xevxpov eaxl xou ABr xuxXou, 
Xai] eaxlv f] ZA xfj Zr. raxXiv, enel xo H ar)[ieTov xevxpov 
eaxl xou AAE xuxXou, iar\ eaxlv f) HA xfj HA. eSeix^ 



at point A. 



H 




Thus, if two circles touch one another internally, [and 
their centers are found], then the straight-line joining 
their centers, [being produced], will fall upon the point 
of union of the circles. (Which is) the very thing it was 
required to show. 

Proposition 12 

If two circles touch one another externally then the 
(straight-line) joining their centers will go through the 
point of union. 

B 




For let two circles, ABC and ADE, touch one an- 
other externally at point A, and let the center F of ABC 
have been found [Prop. 3.1], and (the center) G of ADE 
[Prop. 3.1] . I say that the straight-line joining F to G will 
go through the point of union at A. 

For (if) not then, if possible, let it go like FCDG (in 
the figure), and let AF and AG have been joined. 

Therefore, since point F is the center of circle ABC, 
FA is equal to FC. Again, since point G is the center of 
circle ADE, GA is equal to GD. And FA was also shown 



82 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



Be xal f] ZA xrj ZT lory ai apa ZA, AH xdu; Zr, HA 1'aai 
eto£v woxe okr\ f) ZH xwv ZA, AH ja.e[£cov eaxiv dXXa xal 
eXdxxcov onep eaxlv dSuvaxov. oux apa f] arco xou Z era 
xo H era^euyvujjievr] eGi&e'ia 8id xrjg xaxd xo A eraxcpfjc; oux 
eXeuoexai- Si' aGxrjc; apa. 

'Edv apa 860 xuxXoi ecpdnxiovxai dXXrjXwv exxoc;, f) era 
xd xevxpa auxGv eraCeuyvu^evr] [eO'dela] 81a xrjc; enacprjc 
eXeuaexai - ouep eSei 8eTc;ai. 




EE yap 8uvax6v, xuxXoc; 6 ABrA xuxXou xou EBZA 
ecpanxecrdco upoxepov evxoc; xaxd nXeiova ar)|jieTa fj ev xd A, 
B. 

Kal EiX^cpdco xou ^iev ABrA xuxXou xevxpov xo H, xou 
8e EBZA xo 6. 

; H apa duo xou H era xo © eTu^euyvujievr] era xd B, 
A TteaeTxai. tuttxcxm (be; #j BH6A. xal end xo H a/jjietov 
xevxpov eoxl xou ABrA xuxXou, Tor) eaxlv #j BH xfj HA- 
(lei^tdv apa f) BH xrjc; OA- ttoXXG apa (jiei^wv f] B9 xrjc; OA. 
ndXiv, knzi xo 6 a/jjielov xevxpov eaxl xou EBZA xuxXou, 
Tar) eaxlv f) B6 xfj OA- eBeixii/] Be auxfjg xal ttoXXG jiei^cov 
ojiep douvaxov oux apa xuxXog xuxXou ecpdnxexai evxoc 
xaxd TtXeiova a/jjiela fj ev. 

Aeyco orj, oxi ou6e exxoc;. 

Ei yap Suvaxov, xuxXoc; 6 ArK xuxXou xou ABrA 
ecpaTixeaiJco exxoc; xaxd TtXeiova ar)(iela f] ev xd A, T, xal 
eTie^eu)cdw f\ AT. 

"Enzl ouv xuxXgjv x£>v ABrA, ArK ei'XrjTtxai km. xrjc; 
Ttepicpepeiac; exaxepou 860 xu)(6vxa or)(ieTa xd A, T, f) era 
xd ar)U.eTa era^euyvu^evr) cu^ela evxoc; exaxepou neaelxar 
dXXa xou [iev ABrA evxoc; eneaev, xou Be ArK exxoc;- 
ojiep axoTtov oux apa xuxXoc; xuxXou ecpdnxexai exxoc; xaxd 
TiXebva ar^ela fj ev. eBeiy^r] Be, 6x1 ouBe evxoc;. 



(to be) equal to FC. Thus, the (straight-lines) FA and 
AG are equal to the (straight-lines) FC and GD. So the 
whole of FG is greater than FA and AG. But, (it is) also 
less [Prop. 1.20]. The very thing is impossible. Thus, the 
straight-line joining F to G cannot not go through the 
point of union at A. Thus, (it will go) through it. 

Thus, if two circles touch one another externally then 
the [straight-line] joining their centers will go through 
the point of union. (Which is) the very thing it was re- 
quired to show. 

Proposition 13 

A circle does not touch a(nother) circle at more than 
one point, whether they touch internally or externally 




For, if possible, let circle ABDC 1 * touch circle EBFD — 
first of all, internally — at more than one point, D and B. 

And let the center G of circle ABDC have been found 
[Prop. 3.1], and (the center) H of EBFD [Prop. 3.1]. 

Thus, the (straight-line) joining G and H will fall on 
B and D [Prop. 3.11]. Let it fall like BGHD (in the 
figure) . And since point G is the center of circle ABDC, 
BG is equal to GD. Thus, BG (is) greater than HD. 
Thus, BH (is) much greater than HD. Again, since point 
H is the center of circle EBFD, BH is equal to HD. 
But it was also shown (to be) much greater than it. The 
very thing (is) impossible. Thus, a circle does not touch 
a(nother) circle internally at more than one point. 

So, I say that neither (does it touch) externally (at 
more than one point) . 

For, if possible, let circle ACK touch circle ABDC 
externally at more than one point, A and C. And let AC 
have been joined. 

Therefore, since two points, A and C, have been taken 
at random on the circumference of each of the circles 
ABDC and ACK , the straight-line joining the points will 
fall inside each (circle) [Prop. 3.2]. But, it fell inside 
ABDC, and outside ACK [Def. 3.3]. The very thing 



83 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



KuxXog dpa xuxXou oux ecpditxexai xaxd TtXeiova arj^ieTa 
rj [xotd'] ev, edv xe evxoc edv xe exxoc; ecpdnxrjxai' ojtep e8ei 
8eTc;ai. 



t The Greek text has "ABCD", which is obviously a mistake. 

18'. 

'Ev xuxXcp al faai euifteTai faov djie/ouaiv and xou 
xevxpou, xai al faov dne/ouaai dno xou xevxpou faai 
dXXrjXaic eiaiv. 




A 

"Eaxw xuxXoc; 6 ABrA, xai ev auxco faai eO'delai eaxco- 
aav al AB, EA- Xeyco, oxi al AB, EA faov aTte)(ouaiv arco 
xou xevxpou. 

EiXr)cp , dco yap xo xevxov xou ABrA xuxXou xdi eaxw xo 
E, xai and xou E era xd<; AB, TA xdiJexoi fjx'dwcjav al EZ, 
EH, xdi eirsCeOx'OwCTav al AE, EE 

'Eitei ouv eicdeTd xu; 8la xou xevxpou f] EZ euiDeTdv xiva 
^.f] 8ia xou xevxpou xf)v AB Ttpoc; op'da? xe^ivei, xai 8[)(a 
auxfjv xe^vei. far] apa f] AZ xfj ZB- SixcXfj apa f] AB xrj<; 
AZ. 8id xd auxd Sf] xai f] TA xrj<; TH eaxi SiirArj" xai eaxiv 
far] f) AB xfj EA- far) apa xai f) AZ xfj TH. xai ckci far] eaxiv 
f] AE xfj EE, iaov xai xo d^o xfj<; AE xw duo xrj<; EE dXXd 
tu [i£v arco xrj<; AE laa xd duo xwv AZ, EZ' op'dr) yap f) 
7ip6<; xw Z ywvia- xw 8e diio xfj? Er laa xd diio xwv EH, HE 
op'df] yap f) 7tp6<; iu H y«v(a- xd apa dfto xtov AZ, ZE faa 
eaxi xoTc duo xwv TH, HE, Sv xo duo xrj<; AZ iaov eaxi tu 
drco xfj? TH- far) yap eaxiv f] AZ xfj ITE Xoikov apa xo duo 
xfjc; ZE tw duo xrjc EH faov eaxiv far] apa f] EZ xfj EH. ev 
Se xuxXw iaov dTte^eiv duo xou xevxpou eu-delai Xeyovxai, 
oxav al and xou xevxpou in auxdc; xdiJexoi dyo^tevai faai 
CSaiv al apa AB, EA faov duexouaiv ano xou xevxpou. 

AXXd 8f) al AB, EA eu'delai faov dnexexwaav duo xou 
xevxpou, xouxeaxiv far) eaxco f) EZ xfj EH. Xeyw, oxi far] 
eaxi xai f] AB xfj EA. 



(is) absurd. Thus, a circle does not touch a(nother) circle 
externally at more than one point. And it was shown that 
neither (does it) internally. 

Thus, a circle does not touch a(nother) circle at more 
than one point, whether they touch internally or exter- 
nally. (Which is) the very thing it was required to show. 



Proposition 14 

In a circle, equal straight-lines are equally far from the 
center, and (straight-lines) which are equally far from the 
center are equal to one another. 



B/ \ 
\ E^G 



\ F \ \\ / 
A 

Let ABD0 be a circle, and let AB and CD be equal 
straight-lines within it. I say that AB and CD are equally 
far from the center. 

For let the center of circle ABDC have been found 
[Prop. 3.1], and let it be (at) E. And let EF and EG 
have been drawn from (point) E, perpendicular to AB 
and CD (respectively) [Prop. 1.12]. And let AE and EC 
have been joined. 

Therefore, since some straight-line, EF, through the 
center (of the circle), cuts some (other) straight-line, AB, 
not through the center, at right-angles, it also cuts it in 
half [Prop. 3.3]. Thus, AF (is) equal to FB. Thus, AB 
(is) double AF. So, for the same (reasons), CD is also 
double CG. And AB is equal to CD. Thus, AF (is) 
also equal to CG. And since AE is equal to EC, the 
(square) on AE (is) also equal to the (square) on EC. 
But, the (sum of the squares) on AF and EF (is) equal 
to the (square) on AE. For the angle at F (is) a right- 
angle [Prop. 1.47]. And the (sum of the squares) on EG 
and GC (is) equal to the (square) on EC. For the angle 
at G (is) a right-angle [Prop. 1.47]. Thus, the (sum of 
the squares) on AF and FE is equal to the (sum of the 
squares) on CG and GE, of which the (square) on AF 
is equal to the (square) on CG. For AF is equal to CG. 



84 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



Toiv ydp auxfiv xaxaoxeuaaiDevTWv 6[io[cl><; 8eic;o^.ev, 
oxi SiTtXfj eaxiv f) jiev AB tt^c: AZ, f) 8e TA xfjc; TH- xal end 
Tar) eaxiv f) AE xfj TE, i'aov eaxl xo ano xfjc; AE x« arco 
xfjc; TE' dXXd xco [iev and xfjc; AE Taa eaxl xd duo xc5v EZ, 
ZA, xcp Se (xtco xfjc; TE Taa xd duo xfiv EH, HT. xd dpa arco 
xfiv EZ, ZA foot eaxl xolc; dno x£5v EH, HT- Sv xo arco xfjc; 
EZ xfi duo xfjc; EH eaxiv iaov Tar) yap /) EZ xfj EH' Xoitiov 
dpa xo duo xfjc; AZ i'aov eaxl xw duo xfjc; TH- lar\ dpa f) AZ 
xfj TH- xal eaxi xfjc; jiev AZ SiTtXfj f] AB, xfjc; Se TH SiTtXfj 
f) TA- Tar) dpa f] AB xfj TA. 

'Ev xuxXw dpa al Taai eu-delai I'aov aTtexouaiv duo 
xou xevxpou, xal ai I'oov dnexouaai duo xou xevxpou Taai 
dXXfjXaic; eto£v OTtep e8ei Bel^ai. 



t The Greek text has "ABCD", which is obviously a mistake. 

IS . 

'Ev xuxXcp (ieyiaxT] ^ev f] Bidjiexpoc;, xwv 8e dXXcov del 
f] eyyiov xou xevxpou xfjc; duwxepov ^.si^wv saxiv. 

'Eaxco xuxXoc; 6 ABrA, 8id[iexpoc; 8e auxou eox« #j A A, 
xevxpov 8e xo E, xal eyyiov ^xev xfjc; AA 8ia^texpou eaxo f) 
Br, aTioxepov 8e f\ ZH- Xeyw, oxi [leytaxrj [Lev eoxiv f] AA, 
^ielC«v 8e f] Br xfjc; ZH. 

"H)edtL>aav yap dito xou E xevxpou era xdc; Br, ZH 
xdiftexoi ai E6, EK. xal ensl eyyiov ^tev xou xevxpou eaxiv 
f] Br, dnwxepov 8e f] ZH, ^ei^wv dpa f] EK xfjc; E6. xeirj'do 
xfj E9 lay] f] EA, xal 8id xou A xfj EK Tipoc; op-Quc, d)fdeTaa 
f] AM 8if]X'S« era to N, xal eTte^eux^waav al ME, EN, ZE, 
EH. 

Kal CTtel iat] eaxiv f] E9 xfj EA, lot] eaxl xal f] BT xfj 
MN. TtdXiv, euel iarj eaxiv f] ^xev AE xfj EM, f] 8e EA x^ 
EN, f) dpa AA xalc; ME, EN I'ar) eaxiv. dXX'' ai jiev ME, EN 
xfjc; MN ^.eiCovec; eiaiv [xal f) AA xfjc; MN (iciCwv eaxiv], 
iar] 8e f) MN xfj Br- f) AA dpa xfjc; Br [ie(Cwv eaxiv. xal 
CTtel 8uo ai ME, EN 8uo xalc; ZE, EH i'aai eiaiv, xal ywvla 
f] bub MEN ywviac; xfjc; UTto ZEH ^lei^wv [eaxiv], [3daic; dpa 



Thus, the remaining (square) on FE is equal to the (re- 
maining square) on EG. Thus, EF (is) equal to EG. And 
straight-lines in a circle are said to be equally far from 
the center when perpendicular (straight-lines) which are 
drawn to them from the center are equal [Def. 3.4]. Thus, 
AB and CD are equally far from the center. 

So, let the straight-lines AB and CD be equally far 
from the center. That is to say, let EF be equal to EG. I 
say that AB is also equal to CD. 

For, with the same construction, we can, similarly, 
show that AB is double AF, and CD (double) CG. And 
since AE is equal to CE, the (square) on AE is equal to 
the (square) on CE. But, the (sum of the squares) on 
EF and FA is equal to the (square) on AE [Prop. 1.47]. 
And the (sum of the squares) on EG and GC (is) equal 
to the (square) on CE [Prop. 1.47]. Thus, the (sum of 
the squares) on EF and FA is equal to the (sum of the 
squares) on EG and GC, of which the (square) on EF is 
equal to the (square) on EG. For EF (is) equal to EG. 
Thus, the remaining (square) on AF is equal to the (re- 
maining square) on CG. Thus, AF (is) equal to CG. And 
AB is double AF, and CD double CG. Thus, AB (is) 
equal to CD. 

Thus, in a circle, equal straight-lines are equally far 
from the center, and (straight-lines) which are equally far 
from the center are equal to one another. (Which is) the 
very thing it was required to show. 



Proposition 15 

In a circle, a diameter (is) the greatest (straight-line), 
and for the others, a (straight-line) nearer to the center 
is always greater than one further away. 

Let ABCD be a circle, and let AD be its diameter, 
and E (its) center. And let BC be nearer to the diameter 
AD ,t and FG further away. I say that AD is the greatest 
(straight-line), and BC (is) greater than FG. 

For let EH and EK have been drawn from the cen- 
ter E, at right-angles to BC and FG (respectively) 
[Prop. 1.12]. And since BC is nearer to the center, 
and FG further away, EK (is) thus greater than EH 
[Def. 3.5]. Let EL be made equal to EH [Prop. 1.3]. 
And LM being drawn through L, at right-angles to EK 
[Prop. 1.11], let it have been drawn through to N. And 
let ME, EN, FE, and EG have been joined. 

And since EH is equal to EL, BC is also equal to 
MN [Prop. 3.14]. Again, since AE is equal to EM, and 
ED to EN, AD is thus equal to ME and EN. But, ME 
and EN is greater than MN [Prop. 1.20] [also AD is 



85 



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ELEMENTS BOOK 3 



f) MN pdaecoc; xfjc ZH ^ei^tov eaxlv. dXXd fj MN xfj Br greater than MJV], and MJV (is) equal to SC. Thus, AL> 
sSeixdr] Tar) [xod fj Br xfjc ZH ^ie[£«v eaxlv] . (jlsylcttt) [iev is greater than BC. And since the two (straight-lines) 
dpa fj AA Bid^iexpoc;, liel^tov 8s fj Br xfjc; ZH. ME, EN are equal to the two (straight-lines) EE, EG 

(respectively), and angle MEN [is] greater than angle 
FEG, X the base MN is thus greater than the base FG 
[Prop. 1.24]. But, MN was shown (to be) equal to BC 
[(so) BC is also greater than FG]. Thus, the diameter 
AD (is) the greatest (straight-line), and BC (is) greater 
than FG. 




'Ev xuxXw dpa ^syiaxr) jiev eaxiv f| 8id[iexpoc;, xfiv Be: Thus, in a circle, a diameter (is) the greatest (straight- 
dXXwv del f] syyiov xoO xevxpou xfjc ditwxepov (jeiCuv eaxlv line), and for the others, a (straight-line) nearer to the 
oTtep e8ei Bei^ai. center is always greater than one further away. (Which 

is) the very thing it was required to show. 

f Euclid should have said "to the center", rather than "to the diameter AD", since BC, AD and FG are not necessarily parallel, 
t This is not proved, except by reference to the figure. 



H xfj Sia^expw xou xuxXou 7ip6<; opMc; dm' dxpac; 
dyo^ievr] exxoc; TteaeTxai xou xuxXou, xal sic; xov [iexac;u 
xoitov xfjc; xe eu'delac; xal xfjc; Ttepicpspeiae; exepa eu-dela ou 
TtapejjmeaeTxai, xal fj [lev xou fjuixuxXfou Y wv i a aTtdarjc; 
Y«v(a<; 6e;elac; eu-duYpd^ou ^teiCtov eaxlv, f) 8e XoiTtf) 
eXdxxwv. 

TCaxo xuxXoc; 6 ABT nepl xevxpov xo A xal 8id^texpov 
xfjv AB- Xsy", 8xi fj duo xou A xfj AB Ttpoc; opMc; an' 
dxpac; dYo^ievr] exxoc; neaeTxai xou xuxXou. 

Mf] Ydp, dXX' el 8uvax6v, mnisxw evxoc; wc; f) TA, xal 
£Tie^£U)cd« fj AT. 

'EticI larj eaxlv f] AA xfj AT, ia/j eaxl xal ywvla fj uti:6 
AAr ywvia xfj utto ArA. opiSfj 8s fj utto AAr- opiSfj dpa 
xal fj Otto ArA' xpiytiivou 8f) xou ArA al Buo y<^io(1 oti 
utto AAT, ATA 6uo op-dalc; I'oai eialv oTiep eaxlv dSuvaxov. 
oux dpa f) arco xou A at]y.e'\.ou xfj BA Ttpoc; 6p$a<; aYo^evr) 
evxoc; neaeTxai xou xuxXou. by.oioic, 8fj SeTd;o^ev, oxi ou8' 
stxl xfjc; nspicpspeiac;- exxoc; dpa. 



Proposition 16 

A (straight-line) drawn at right-angles to the diameter 
of a circle, from its end, will fall outside the circle. And 
another straight-line cannot be inserted into the space be- 
tween the (aforementioned) straight-line and the circum- 
ference. And the angle of the semi-circle is greater than 
any acute rectilinear angle whatsoever, and the remain- 
ing (angle is) less (than any acute rectilinear angle) . 

Let ABC be a circle around the center D and the di- 
ameter AB. I say that the (straight-line) drawn from A, 
at right-angles to AB [Prop 1.11], from its end, will fall 
outside the circle. 

For (if) not then, if possible, let it fall inside, like CA 
(in the figure), and let DC have been joined. 

Since DA is equal to DC, angle DAG is also equal 
to angle ACD [Prop. 1.5]. And DAG (is) a right-angle. 
Thus, ACD (is) also a right-angle. So, in triangle ACD, 
the two angles DAG and ACD are equal to two right- 
angles. The very thing is impossible [Prop. 1.17]. Thus, 
the (straight-line) drawn from point A, at right-angles 



8G 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



B 




IIiTcxexco d>c f] AE- Xeyio 8r], oxi eic xov jiexacu xotiov 
xfjc xe AE eu'deiac xal xfjc T9A Txeptcpepeta^ exepa eu-dela 
ou Tiape^iTceaeTxai. 

Et yap 8uvax6v, Tiape|iTiiTixex6) 6? f] ZA, xal fjx'dco &ti6 
xoO A ar](!£iou era. xfjv ZA xdcdsxcx; f) AH. xal CTiel op-df] 
eaxiv f) utio AHA, eXdxxtov 8e op'dfjc f] utio AAH, ^teii^v 
apa f) A A xfjc AH. iar) Be f] A A xfj A6- (j.£l^cov apa f] A0 
xfjc AH, f] eXdxxwv xfjc ^el^ovoc oTiep eaxlv dBuvaxov. oux 
apa etc xov liexacu xotcov xfjc xe eMeiac xal xfjc Tiepicpepelac 
exepa eu-dela Tiape^iTceaeTxai. 

Aeyco, oxi xal f) \±ev xoO fjuixuxXlou ycovia f] Tiepiexo^ievr] 
utio xe xfjc BA eu'deiac xal xfjc T0A Tiepicpepelac aatdarjc 
yoviac o^etac eu-duypd^tuou ^iei£«v eaxiv, f] Be XoiTif] f] tic- 
pieXO[Levf] utio xe xfjc T9A Tiepicpepelac xal xfjc AE eu-Mac 
a7idar]c ywvlac o^eiac eu-duypd^uou eXdxxwv eaxlv. 

EE yap eaxl xic ywvia euiJuypa^oc [lz'\Z,(J\> [lev xfjc 
7iepiexo[ievr)c Otio xe xfjc BA eu'deiac xal xfjc T9A Tiepi- 
cpepelac, eXdxxtov Be xfjc Tiepiexo[ievr)c utio xe xfjc T0A 
Tiepicpepelac xal xf)c AE eu'deiac, eic xov ^texacu xotiov xfjc 
xe rOA Tiepicpepelac xal xfjc AE cu^elac euiJeTa Tiapeu- 
TieaeTxai, fjxic Tioifjaei jiel^ova [lev xfjc 7iepie)(o^evr)c Otio 
xe xfjc BA eu'deiac xal xfjc TQA Tiepicpepelac utio eu'deifiv 
7iepiexo[ievr)v, eXdxxova Be xfjc Tiepiexo^ievrjc Otco xe xfjc 
rOA Tiepicpepelac xal xfjc AE eu'deiac. ou Tiape^miTixei Be- 
oux apa xfjc Tiepiexo^ev7)c ywvlac utio xe xfjc BA eu'deiac 
xal xfjc rOA Tiepicpepelac eaxai fiel^cov 6$eTa utio eu'deifiv 
Tiepiexo^evT), ouBe [jurjv eXdxxwv xfjc Tiepiexo^ievr]c Otco xe 
xfjc T0A Tiepicpepelac xal xfjc AE eu'deiac. 



to BA, will not fall inside the circle. So, similarly, we 
can show that neither (will it fall) on the circumference. 
Thus, (it will fall) outside (the circle) . 



B 




Let it fall like AE (in the figure) . So, I say that another 
straight-line cannot be inserted into the space between 
the straight-line AE and the circumference CHA. 

For, if possible, let it be inserted like FA (in the fig- 
ure), and let DG have been drawn from point D, perpen- 
dicular to FA [Prop. 1.12]. And since AGD is a right- 
angle, and DAG (is) less than a right-angle, AD (is) 
thus greater than DG [Prop. 1.19]. And DA (is) equal 
to DH. Thus, DH (is) greater than DG, the lesser than 
the greater. The very thing is impossible. Thus, another 
straight-line cannot be inserted into the space between 
the straight-line (AE) and the circumference. 

And I also say that the semi-circular angle contained 
by the straight-line BA and the circumference CHA is 
greater than any acute rectilinear angle whatsoever, and 
the remaining (angle) contained by the circumference 
CHA and the straight-line AE is less than any acute rec- 
tilinear angle whatsoever. 

For if any rectilinear angle is greater than the (an- 
gle) contained by the straight-line BA and the circum- 
ference CHA, or less than the (angle) contained by the 
circumference CHA and the straight-line AE, then a 
straight-line can be inserted into the space between the 
circumference CHA and the straight-line AE — anything 
which will make (an angle) contained by straight-lines 
greater than the angle contained by the straight-line BA 
and the circumference CHA, or less than the (angle) 
contained by the circumference CHA and the straight- 
line AE. But (such a straight-line) cannot be inserted. 
Thus, an acute (angle) contained by straight-lines cannot 
be greater than the angle contained by the straight-line 
BA and the circumference CHA, neither (can it be) less 
than the (angle) contained by the circumference CHA 
and the straight-line AE. 



87 



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I16pia[Jia. 

'Ex 6r] xouxou cpavepov, oxi f] xrj Bia^iexpw toO xuxXou 
npoq op'&a.z dm' axpa<; &yo(ji£vr] ecpdrcxexai xou xuxXou 
[xai oxi eO'deTa xuxXou xa$' ev ^xovov scpdrcxexai ar^eiov, 
£Tiei.8r]Txep xal f] xaxa 5uo auxo au^pdXXouaa svxoi; auxoO 
TUTtxouaa eSsi/iSr]]- onep eBsi SeT^ai. 



Arco xou hoftevToc, ay][ieiou xoO Scdsvxoi; xuxXou scpa- 
uxo^ievTjv euiiteTav ypa^^v dtYaysTv. 

A 




"Eaxco xo ^jlsv Bo'dev ar)ueiov xo A, 6 8e Bo'dek; xuxXo<; 
6 BrA- 8a Sr) aito xou A ar^dou xoO BrA xuxXou ecpa- 
7ixo[ievr)v eu'deTocv Ypa^rjv dyayETv. 

ElX^cp-dw y^P TO xevxpov xoO xuxXou xo E, xai 
imZ^X^ ^ AE, xdi xsvxpw [Lev xw E Siaaxrj^taxi 8s tu 
EA xuxXo<; yeypdcp'dG) 6 AZH, xal duo xou A xfj EA 7tp6<; 
opMc fj)cdo ^ AZ, xal ensCeux'dwaav ad EZ, AB- Xey«, 
oxi dmo xou A arjudou xou BrA xuxXou eqxntxouevr] fjxxai 
f) AB. 

'Etcei yap xo E xevxpov eaxl xwv BrA, AZH xuxXwv, 
Xar] apa eaxlv f] ^isv EA xrj EZ, f] 5e EA xrj EB- Suo 8r) 
ai AE, EB 8uo xau; ZE, EA taai riaiv xal Y^viav xoivrjv 
7i£pi£)(ouai xf|v 7tp6<; xw E- pdaic; apa f\ AZ pdaei xrj AB 
Xai] saxiv, xal xo AEZ xpiYWvov x£> EBA xpiY^vw laov 
eaxiv, xal ai Xomal Y^wiai xalc; Xoircau; Y^viaic;- Xor\ apa r\ 
bub EAZ xrj Otto EBA. opfir} 8e f) Otto EAZ- op-dr) apa xal r) 
U7i6 EBA. xa[ eaxiv f) EB ex xou xevxpou- f) 8e xrj Bia^texpw 
xou xuxXou Ttp6<; 6pM<; arc' axpac; ayo[Levr] ecpdnxexai xou 
xuxXou- f) AB apa ecpaTtxexai xou BrA xuxXou. 

Atco xou apa So'devxoc; ar)[ieiou xou A xou So-devxoc; 
xuxXou xou BrA sqjaKxo^svr] eui9sTa Ypa^r] fjxxai f) AB- 
onep ebei Ttoifjaai. 



Corollary 

So, from this, (it is) manifest that a (straight-line) 
drawn at right-angles to the diameter of a circle, from 
its extremity, touches the circle [and that the straight-line 
touches the circle at a single point, inasmuch as it was 
also shown that a (straight-line) meeting (the circle) at 
two (points) falls inside it [Prop. 3.2] ]. (Which is) the 
very thing it was required to show. 

Proposition 17 

To draw a straight-line touching a given circle from a 
given point. 

A 




Let A be the given point, and BCD the given circle. 
So it is required to draw a straight-line touching circle 
BCD from point A. 

For let the center E of the circle have been found 
[Prop. 3.1], and let AE have been joined. And let (the 
circle) AFG have been drawn with center E and radius 
EA. And let DF have been drawn from from (point) D, 
at right-angles to EA [Prop. 1.11]. And let EF and AB 
have been joined. I say that the (straight-line) AB has 
been drawn from point A touching circle BCD. 

For since E is the center of circles BCD and AFG, 
EA is thus equal to EF, and ED to EB. So the two 
(straight-lines) AE, EB are equal to the two (straight- 
lines) FE, ED (respectively). And they contain a com- 
mon angle at E. Thus, the base DF is equal to the 
base AB, and triangle DEF is equal to triangle EBA, 
and the remaining angles (are equal) to the (corre- 
sponding) remaining angles [Prop. 1.4]. Thus, (angle) 
EDF (is) equal to EBA. And EDF (is) a right-angle. 
Thus, EBA (is) also a right-angle. And EB is a ra- 
dius. And a (straight-line) drawn at right-angles to the 
diameter of a circle, from its extremity, touches the circle 
[Prop. 3.16 corr.]. Thus, AB touches circle BCD. 

Thus, the straight-line AB has been drawn touching 



88 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



IT]'. 

'Edv xuxXou ecpdTCxrjxai tic; eu'deTa, duo 8s xou xevxpou 
era xr)v 6tcpr]v era£eu)edfi ECcdeia, #] eraCeux'delaa xd-dexoc; 
eaxai era xrjv ecpa7ixouevr)v. 




KuxXou ydp xou ABr ecpaTtxecrda) xlc eO^eTa f\ AE xaxd 
xo T ar)ueTov, xdi eiXr|cp , d« xo xevxpov xou ABr xuxXou xo 
Z, xal duo xou Z era xo T C7teCeux'0c>) f) ZE Xey«, oxi #] Zr 
xd^exoc; eaxiv era x/]v AE. 

El ydp [ir\, r\x&^> duo xou Z era xrjv AE xd-dexoc; f\ ZH. 

'Eitel oGv f) Grab ZHr ywvia op'drj eaxiv, o^ela dpa eaxlv 
f] (mo ZrH- (mo 5e xr|v [ieiCova yaviav f] ^eiX«v uXeupd 
(moxeivei- ueiCwv dpa f] Zr xfj? ZH- lor\ 5e f] Zr xfj ZB- 
[ie'i^v dpa xal f] ZB xfjc ZH f] eXdxxwv xrjc; [lei^ovoc;- oracp 
eaxlv dBuvaxov. oux dpa f] ZH xd-dexoc; eaxiv era xr|v AE. 
o^ioiwg 5r] SeT^ouev, oxi ou8' aXXr] xi? kX/]v xrjc ZE f] Zr 
dpa xd-dexoc; eaxiv era x/]v AE. 

'Eav dpa xuxXou ecpaTixrjxai tic, euiJeTa, drab Be xou 
xevxpou era x/]v dcprjv eraCeux'dfj xic; eu-dela, f] eraCeux'deTaa 
xd-dexoi; eaxai era x/]v ecpaTtxo^ievr]v oraep e5ei BeT^ai. 



'Eav xuxXou ecpdn;xr|xai xu; euiSeTa, drab 8e xrjc; dcpfjc; xfj 
ecpanxoL/evr] 7tp6<; op'ddi; [ycoviac;] eu^eTa yp°W^ &x®% £™- 
xrjc; dx^eiorjc; eaxai xo xevxpov xou xuxXou. 

KuxXou yap xou ABr ecpararecrdco xlc; eui&eTa f] AE xaxd 
xo T arj^ieTov, xal drab xou T xfj AE 7tp6<; op-ddc; f]x'd« r) 
IA- Xeyco, oxi era xrjc Ar eaxi xo xevxpov xou xuxXou. 



the given circle BCD from the given point A. (Which is) 
the very thing it was required to do. 

Proposition 18 

If some straight-line touches a circle, and some 
(other) straight-line is joined from the center (of the cir- 
cle) to the point of contact, then the (straight-line) so 
joined will be perpendicular to the tangent. 




D 



For let some straight-line DE touch the circle ABC at 
point C, and let the center F of circle ABC have been 
found [Prop. 3.1], and let FC have been joined from F 
to C. I say that FC is perpendicular to DE. 

For if not, let FG have been drawn from F, perpen- 
dicular to DE [Prop. 1.12]. 

Therefore, since angle FGC is a right-angle, (angle) 
FCC is thus acute [Prop. 1.17]. And the greater angle is 
subtended by the greater side [Prop. 1.19]. Thus, FC (is) 
greater than FG. And FC (is) equal to FB. Thus, FB 
(is) also greater than FG, the lesser than the greater. The 
very thing is impossible. Thus, FG is not perpendicular to 
DE. So, similarly, we can show that neither (is) any other 
(straight-line) except FC. Thus, FC is perpendicular to 
DE. 

Thus, if some straight-line touches a circle, and some 
(other) straight-line is joined from the center (of the cir- 
cle) to the point of contact, then the (straight-line) so 
joined will be perpendicular to the tangent. (Which is) 
the very thing it was required to show. 

Proposition 19 

If some straight-line touches a circle, and a straight- 
line is drawn from the point of contact, at right- [angles] 
to the tangent, then the center (of the circle) will be on 
the (straight-line) so drawn. 

For let some straight-line DE touch the circle ABC at 
point C. And let CA have been drawn from C, at right- 



89 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



angles to DE [Prop, 
circle is on AC. 



1.11]. I say that the center of the 




Mr] ydp, dXX' et Suvaxov, eaxto to Z, xal sne^sux'dco f\ 

rz. 

'EticI [ouv] xuxXou toO ABr ecpdiixexai tic; euiMa f] AE, 
duo 6e tou xevxpou era x/]v acp/jv CTiei^euxxai f] Zr, f] Zr dpa 
xd-dexoc; eaxiv enl xrjv AE- op'&f] dpa eaxlv f) utio ZrE. eaxi 
6e xdi #j utio ArE op-dry Tor) dpa eaxlv f) Otto ZrE xfj utio 
ArE #j eXdxxwv xrj jiei^ovi- oTiep eaxlv dBuvaxov. oux dpa 
xo Z xevxpov eaxi xoO ABr xuxXou. o^toioc; 8f) 5e(c;o^ev, 
oxi o08' dXXo xi TtXrjv era xrjc Ar. 

Edv dpa xuxXou ecpaTixr]xa( tic, eu'deTa, duo 5e xfjc; acpfjc; 
xfj ecpa7ixo[ievr) Ttpoc; op'ddc; euiDeTa ypa^f) d)fdrj, era xfjc 
d)(Tf)eiar)c; eaxai xo xevxpov xou xuxXou- ouep e8ei 8eTc;ai. 




For (if) not, if possible, let F be (the center of the 
circle), and let CF have been joined. 

[Therefore], since some straight-line DE touches the 
circle ABC, and FC has been joined from the center to 
the point of contact, FC is thus perpendicular to DE 
[Prop. 3.18]. Thus, FCE is a right-angle. And ACE 
is also a right-angle. Thus, FCE is equal to ACE, the 
lesser to the greater. The very thing is impossible. Thus, 
F is not the center of circle ABC. So, similarly, we can 
show that neither is any (point) other (than one) on AC. 

Thus, if some straight-line touches a circle, and a straight- 
line is drawn from the point of contact, at right-angles to 
the tangent, then the center (of the circle) will be on the 
(straight-line) so drawn. (Which is) the very thing it was 
required to show. 



x . 

*Ev xuxXtp f] Tipoc; xfi xevxpw yiovia SiTiXaaitov eaxi xfjc; 
Ttpoc; xrj Ttepicpepeia, oxav xrjv auxfjv Tiepicpepeiav pdaiv ex«- 
aiv od ycoviai. 

'Eaxco xuxXoc; 6 ABr, xal Tipoc; (iev xw xevxpcp auxou 
ywvia eaxw f] utio BET, Tipoc; 8e xfj Ttepicpepeia f] utio BAT, 
exexoaav 8e xf|V auxrjv Tiepicpepeiav pdaiv xrjv Br- Xeyto, 
oxi SircXacuov eaxlv f) Gtio BET ywvia xfjc; utio BAr. 

ETiiCeux'delaa yap f] AE Sifix'O" era to Z. 

'EticI ouv iar] eaxlv f) EA xfj EB, lay) xal ywvia f) utio 
EAB xfj utio EBA- ai dpa utio EAB, EBA ycoviai xfjc; utio 
EAB BiTiXaaiouc; eiaiv. Xor\ 8e f] utio BEZ xau; utio EAB, 
EBA- xal f] utio BEZ dpa xfjc; utio EAB eaxi 8iTiXfj. Sid xd 
auxd 8r] xal f] utio ZEE xfjc utio EAT eaxi SiTiXfj. 6Xrj dpa 
f] utio BEr bXf]z xfjc; utio BAT eaxi SiTiXfj. 



Proposition 20 

In a circle, the angle at the center is double that at the 
circumference, when the angles have the same circumfer- 
ence base. 

Let ABC be a circle, and let BEC be an angle at its 
center, and BAG (one) at (its) circumference. And let 
them have the same circumference base BC. I say that 
angle BEC is double (angle) BAG. 

For being joined, let AE have been drawn through to 

F. 

Therefore, since EA is equal to EB, angle EAB (is) 
also equal to EBA [Prop. 1.5]. Thus, angle EAB and 
EBA is double (angle) EAB. And BEF (is) equal to 
EAB and EBA [Prop. 1.32]. Thus, BEF is also double 

EAB. So, for the same (reasons), FEC is also double 

EAC. Thus, the whole (angle) BEC is double the whole 
(angle) BAG. 



90 



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ELEMENTS BOOK 3 




r 



B 

KexXdcrdo) 5r] TidXiv, xai eaxco exepa ywvia f] Otto BAr, 
xai SKi^eux'Se'iCTa #] AE expepXf]a , dco era to H. 6[ioi«<; 8f) 
8ei^o[i£v, oxi BiTiXfj eaxiv f) utto HEr ycovia xfjc; utto EAr, 
fiv f) Otto HEB SiTcXfj eaxi xfjc; utto EAB- Xomf) apa f) Otto 
BEr SiTtXfj saxi xfjc; utto BAr. 

'Ev xuxXw apa f\ upoc; xw xevxpa) ycovia SmXaaiwv saxi 
xfjc; Tipoc; xfj Tcepicpepda, oxav xfjv auxf]v Ttepicpepeiav pdaiv 
S)(coaiv [ai ywviai]- oicsp eSei 8eTc;ai. 

xa'. 

'Ev xuxXip ai ev xO auxcp x[if]|jiaxi ywviai iaai dXXfjXaic; 
eiaiv. 




Tiaxw xuxXoc; 6 ABrA, xal ev xw auxfi x^uaxi iu 
BAEA yioviai eaxwaav ai Otto BAA, BEA- Xeyw, oxi ai 
Otto BAA, BEA ytoviai loan dXXfjXaic; eiaiv. 

EiXr]cp'd« yap xou ABrA xuxXou xo xevxpov, xal eaxco 
xo Z, xai eTteCeuyjdwaav ai BZ, ZA. 

Kai eicsi f) [lev Otto BZA yiovia Ttpoc; xai xevxpa> eaxiv, f) 
8e Otto BAA Ttpoc; xfj Ttepicpepeia, xai exouai T1 ^ v aUT ^ v Tie- 
picpepeiav pdaiv xfjv BrA, f] apa utto BZA ya>via SiTtXaaiwv 
eaxi xfjc; Otto BAA. 8ia xa auxa 8f) f) utto BZA xai xfjc; utto 




c 



B 



So let another (straight-line) have been inflected, and 
let there be another angle, BDC. And DE being joined, 
let it have been produced to G. So, similarly, we can show 
that angle GEC is double EDC, of which GEB is double 
EDB. Thus, the remaining (angle) BEC is double the 
(remaining angle) BDC. 

Thus, in a circle, the angle at the center is double that 
at the circumference, when [the angles] have the same 
circumference base. (Which is) the very thing it was re- 
quired to show. 

Proposition 21 

In a circle, angles in the same segment are equal to 
one another. 




Let ABCD be a circle, and let BAD and BED be 
angles in the same segment BAED. I say that angles 
BAD and BED are equal to one another. 

For let the center of circle ABCD have been found 
[Prop. 3.1], and let it be (at point) F . And let BF and 
FD have been joined. 

And since angle BFD is at the center, and BAD at 
the circumference, and they have the same circumference 
base BCD, angle BFD is thus double BAD [Prop. 3.20]. 



91 



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ELEMENTS BOOK 3 



BE A eaxi 8mXot<ov Xar\ dpa f) utio BAA xfj utio BE A. 

'Ev xuxXtp apa al ev xfi auxw x^yjuaxi ytovtcxi i'aai 
dXXr]Xaic eta£v OTiep eSei 8el^ai. 




'Eaxco xuxXo<; 6 ABrA, xal ev auxw xexpdiiXeupov eaxo 
xo ABrA- Xeyw, oxi at diievavxiov yoviai 8uaiv op'dau; i'aai 
eiaiv. 

TkeCeux'dwaav ai Ar, BA. 

'Etiei ouv navxoc; xpiycovou ai xpeu; ycoviai 8uaiv opiDaTt; 
laai eiaiv, xoO ABr apa xpiywvou ai xpeu; ycoviai ai utio 
TAB, ABr, BrA 8uaiv opiJau; i'aai eiaiv. Xor\ 8s t\ fiev utio 
TAB xfj uko BAr- ev yap ic5 auxw xjir^axi eiai xo BAAT- 
f] 8s utio ATB xfj utio AAB- ev yap xc5 auxo x^xr^axi eiai 
xo AArB- 6Xr) apa f] utio AAr xau; Otio BAr, ArB 'ior\ 
eaxiv. xoivf) Tipoaxeio^w f) utio ABr- ai apa utio ABr, 
BAr, ArB xau; utio ABr, AAr i'aai eiaiv. dXX'' ai utio 
ABr, BAr, ArB Suaiv bpftdXc, i'aai eiaiv. xai ai utio ABr, 
AAr dpa 8uaiv op'daic; laai eiaiv. b\ioicdz 8r) Sei^o^tev, oxi 
xai ai utio BAA, ArB ywviai Suaiv opiJau; i'aai eiaiv. 

Ttov apa ev xou; xuxXou; xexpaTiXeupwv ai diievavxiov 
ycoviai 8uaiv op'ddu; laai eiaiv oTiep e8ei Sel^ai. 



xy'. 

'Era xfjc auxfjc eMeiac 8uo x|ir|jiaxa xuxXcov o^ioia xai 
dviaa ou auaxaiDrpexai era xd auxa {iepr\. 

Ei yap 8uvax6v, era xrj<; auxrj<; euTDeia<; xfj<; AB Suo 
x^ir][iaxa xuxXwv ojioia xai dviaa auveaxdxw era xd auxa 
[iepr] xd ArB, AAB, xai 8ir])fdo f] ATA, xai eTre^eu)cd«aav 



So, for the same (reasons), BFD is also double BED. 
Thus, BAD (is) equal to BED. 

Thus, in a circle, angles in the same segment are equal 
to one another. (Which is) the very thing it was required 
to show. 

Proposition 22 

For quadrilaterals within circles, the (sum of the) op- 
posite angles is equal to two right-angles. 




Let ABCD be a circle, and let ABCD be a quadrilat- 
eral within it. I say that the (sum of the) opposite angles 
is equal to two right-angles. 

Let AC and BD have been joined. 

Therefore, since the three angles of any triangle are 
equal to two right-angles [Prop. 1.32], the three angles 
CAB, ABC, and BCA of triangle ABC are thus equal 
to two right-angles. And CAB (is) equal to BDC. For 
they are in the same segment BADC [Prop. 3.21]. And 
ACB (is equal) to ADB. For they are in the same seg- 
ment ADCB [Prop. 3.21]. Thus, the whole of ADC is 
equal to BAC and ACB. Let ABC have been added to 
both. Thus, ABC, BAC, and ACB are equal to ABC 
and ADC. But, ABC, BAC, and ACB are equal to two 
right-angles. Thus, ABC and ADC are also equal to two 
right-angles. Similarly, we can show that angles BAD 
and DCB are also equal to two right-angles. 

Thus, for quadrilaterals within circles, the (sum of 
the) opposite angles is equal to two right-angles. (Which 
is) the very thing it was required to show. 

Proposition 23 

Two similar and unequal segments of circles cannot be 
constructed on the same side of the same straight-line. 

For, if possible, let the two similar and unequal seg- 
ments of circles, ACB and ADB, have been constructed 
on the same side of the same straight-line AB. And let 



92 



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ELEMENTS BOOK 3 



ai TB, AB. 




'Ercei ouv 6\loi6v eaxi to ArB xji/jjia iu AAB x^^axi, 
ojjioia 8e xjir^iaxa xuxAiov eaxl xd 8e)(6[ieva y^viac; laac, 
Tar) apa eaxlv f) uno ArB ycovia xfj Otio AAB f) exxoc; xrj 
evxoc;- onep eaxlv d8uvaxov. 

Oux apa em xrjc; auxrjc; euiSeiac; 860 x|ir][iaxa xuxXwv 
o|ioia xal aviaa auaxai3r]0£xai era xa auxd y.epi]- onep eSei 
8eTc;ai. 



^4CZ? have been drawn through (the segments), and let 
CB and DB have been joined. 




A B 

Therefore, since segment ACB is similar to segment 
ADB, and similar segments of circles are those accept- 
ing equal angles [Def. 3.11], angle ACB is thus equal 
to ADB, the external to the internal. The very thing is 
impossible [Prop. 1.16]. 

Thus, two similar and unequal segments of circles 
cannot be constructed on the same side of the same 
straight-line. 



x5'. 



Proposition 24 



Td em '(awv euiJeiMv o|ioia x^uaxa xuXgjv I'oa aXki]ko\.c, 
eaxiv. 




Similar segments of circles on equal straight-lines are 
equal to one another. 

E 





r A 

"Eaxwaav yap era lawv eu-deiwv x£Sv AB, TA o^ioia 
x^tr]uaxa xuxXwv xa AEB, TZA- Xeyw, oxi I'aov eaxl xo 
AEB x^ifjjia t& TZA x[if]^axi. 

'Ecpap|jio^o|jievou yap xou AEB x^tr]uaxo<; era xo TZA xal 
xidejievou xou [ie\ A o/jjieiou era xo T xrjc; 8e AB euiSetai; 
era xr]v TA, ecpapjioaei xal xo B o/jjielov era xo A arj^ielov 
Sid xo Xat]\i elvai xrjv AB xfj TA- xfjc 8e AB era xrjv TA ecpap- 
y.oc6>ar\c, ecpap^iooei xal xo AEB xjifj^ia era xo TZA. ei yap 
f) AB eu'dela era xrjv TA ecpap^xoaei, xo 8e AEB xjrrj^a era 
xo TZA jit) ecpapjiooei, fjxoi evxoc; auxou neaelxai f\ exxoc; 
f] n:apaXXdc;ei, d>c; xo THA, xal xuxXoc; xuxXov xejivei xaxd 
TiXeiova cnqjiela fj 8uo' onep eaxiv dBuvaxov. oux apa ecpap- 
[ioZ,oy.evf]z xrjc; AB eO'deiac; era xrjv TA oux ecpapjiooei xal 




C D 

For let AEB and CFD be similar segments of circles 
on the equal straight-lines AB and CD (respectively) . I 
say that segment AEB is equal to segment CFD. 

For if the segment AEB is applied to the segment 
CFD, and point A is placed on (point) C, and the 
straight-line AB on CD, then point B will also coincide 
with point D, on account of AB being equal to CD. And 
if AB coincides with CD then the segment AEB will also 
coincide with CFD. For if the straight-line AB coincides 
with CD, and the segment AEB does not coincide with 
CFD, then it will surely either fall inside it, outside (it),^ 
or it will miss like CCD (in the figure), and a circle (will) 
cut (another) circle at more than two points. The very 



93 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



to AEB Turj^ia era to TZA- ecpapjioaei apa, xal taov ai)T« 
eaTai. 

Td apa era lacov eO'deioov o^ioia T^f|^aTa xuxXwv laa 
dXXyjXoic; ecrav oTcep eBei 5eT?ai. 



t Both this possibility, and the previous one, are precluded by Prop. 3.23. 

xs'. 

KuxXou x^jiaxoc; So-devxoc; Tcpoaavaypdcj;ai tov xuxXov, 
ounep eaxi Tjxfj^a. 




r r r 



"Eotw to 5oi9ev xurjpc xuxXou to ABE 8el Bf] toO ABr 
xp^axoc; Ttpoaavaypdcjjai tov xuxXov, ouTcep eaxi xjirjjia. 

Tex[i rjcrdco yap f] Ar Stya xaxd to A, xal rjx'dco duo xou 
A arjueiou xfj Ar Tcpoc; opMc; f] AB, xal erav^eux'&co i) AB- 
f] Otco ABA yovla apa xrjc; Otto BAA f]xoi ^elc^cov eaxlv rj 
tar] fj eXdxxwv. 

"Eaxco TcpoTepov iiei^cov, xal auveoxdxo repot; xfj BA 
eO'dela xal x£> Tcpoc; aOxfj ar^eio tw A Tfj Otco ABA ywvia 
Xor\ f) Otco BAE, xal 8ir])fdw f] AB era to E, xal erceCeux'dco 
f] Er. excel ouv lay] eaxlv rj Otco ABE ycovia xfj Otco BAE, 
lay] apa eaxi xal y) EB eu-dela Tfj EA. xal ercel iar] eaxlv f) 
AA Tfj Ar, xoivf] Se r) AE, Suo Bf] al AA, AE Buo Talc; 
TA, AE laai eialv exaxepa exaxepa- xal ycovia y) Otco AAE 
ycovia xfj Otco TAE eaxiv car)" dp'df) yap exaTepa' pdaic; apa 
f] AE pdoei Tfj TE eaxiv car). dXXd y) AE Tfj BE eSd^r] 
lay]- xal f) BE apa Tfj TE eaxiv Iar)- ai xpelc; apa ai AE, EB, 
Er I'aai dXXf|Xaic; eia(v 6 apa xevxpaS x£> E 8iaaxf]^axi Be 
evl twv AE, EB, Er xuxXoc; ypacpouevoc; f^ei xal Sid xwv 
Xoitcwv ar)[ieiwv xal eaxai Tcpoaavayeypa^evoc;. xuxXou 
apa x^if^axoc; So'devxoc; TtpoaavayeypaTcxai 6 xuxXog. xal 
8fjXov, cbc; to ABr T^ifjiia eXaxxov eaxiv f]^ixuxX(ou Sid to 
to E xevxpov 6xt6c; auxou Tuy)(dveiv. 

'Oiioiwc; [8e] xav f) f) Otco ABA ycovia lay] Tfj Otco BAA, 
xfjc; AA larf, yevo|ievr)<; exaTepa xoov BA, Ar ai Tpelc; al 
AA, AB, Ar Taai dXXfjXaic; eaovxai, xal eaxai to A xevxpov 
toO TcpoaavaTceTcXr)pw^ievou xuxXou, xal SrjXaSf) eaxai to 
ABr fjuixuxXiov. 

'Edv Be f) Otco ABA eXdxxwv fj xfjc; Otco BAA, xal au- 
axrjao^cda Tcpoc; Tfj BA euiJeia xal iS Tcpoc; auxfj ar^elw 



thing is impossible [Prop. 3.10]. Thus, if the straight-line 
AB is applied to CD, the segment AEB cannot not also 
coincide with CFD. Thus, it will coincide, and will be 
equal to it [C.N. 4]. 

Thus, similar segments of circles on equal straight- 
lines are equal to one another. (Which is) the very thing 
it was required to show. 



Proposition 25 



For a given segment of a circle, to complete the circle, 
the very one of which it is a segment. 

A A . A 




Let ABC be the given segment of a circle. So it is re- 
quired to complete the circle for segment ABC, the very 
one of which it is a segment. 

For let AC have been cut in half at (point) D 
[Prop. 1.10], and let DB have been drawn from point 
D, at right-angles to AC [Prop. 1.11]. And let AB have 
been joined. Thus, angle ABD is surely either greater 
than, equal to, or less than (angle) BAD. 

First of all, let it be greater. And let (angle) BAE, 
equal to angle ABD, have been constructed on the 
straight-line BA, at the point A on it [Prop. 1.23]. And 
let DB have been drawn through to E, and let EC have 
been joined. Therefore, since angle ABE is equal to 
BAE, the straight-line EB is thus also equal to EA 
[Prop. 1.6]. And since AD is equal to DC, and DE (is) 
common, the two (straight-lines) AD, DE are equal to 
the two (straight-lines) CD, DE, respectively. And angle 
ADE is equal to angle CDE. For each (is) a right-angle. 
Thus, the base AE is equal to the base CE [Prop. 1.4]. 
But, AE was shown (to be) equal to BE. Thus, BE is 
also equal to CE. Thus, the three (straight-lines) AE, 
EB, and EC are equal to one another. Thus, if a cir- 
cle is drawn with center E, and radius one of AE, EB, 
or EC, it will also go through the remaining points (of 
the segment), and the (associated circle) will have been 
completed [Prop. 3.9]. Thus, a circle has been completed 
from the given segment of a circle. And (it is) clear that 
the segment ABC is less than a semi-circle, because the 
center E happens to lie outside it. 



94 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



tu A xfj Grab ABA yovia larjv, evxo<; xou ABr x^rjpcxoc; 
TteaeTxai xo xevxpov era xrjg AB, xal eaxai SrjXaBr] xo ABr 
xjji/j^ia ^iel£ov fjjiixuxXiou. 

KuxXou apa x^fj^axoc; 8oiL)evxo<; TtpoaavayeypaTixai ° 
xuxXog- onep eSei Hoifjaai. 



[And] , similarly, even if angle ABD is equal to BAD, 
(since) AD becomes equal to each of BD [Prop. 1.6] and 
DC, the three (straight-lines) DA, DB, and DC will be 
equal to one another. And point D will be the center 
of the completed circle. And ABC will manifestly be a 
semi-circle. 

And if ABD is less than BAD, and we construct (an- 
gle BAE), equal to angle ABD, on the straight-line BA, 
at the point A on it [Prop. 1.23], then the center will fall 
on DB, inside the segment ABC. And segment ABC will 
manifestly be greater than a semi-circle. 

Thus, a circle has been completed from the given seg- 
ment of a circle. (Which is) the very thing it was required 
to do. 



X9 



Proposition 26 



'Ev xolc; iaoiz xuxXou; ai Xaai ycoviai era i'awv Ttepicpe- 
peiwv pcprjxaaiv, eav xe 7ip6<; xoTc, xevxpou; eav xe npbq 
xau; Ttspicpepdocu; Sai pepr]X\iiai. 

A. 





K A 

"Eaxcoaav Taoi xuxXoi ol ABr, AEZ xal ev auxou; Taai 
ycoviai eoxwaav upoc; \xk\ xou; xevxpou; ai bub BHr, E9Z, 
Ttpoc; Se xau; Ttepicpepeiau; ai Gtio BAr, EAZ- Xeyco, oxi Tar) 
eaxlv f| BKr rcepicpepeia xfj EAZ nepicpepeia. 

'Erac^e6)cd«oav yap ai Br, EZ. 

Kai CTiel Taoi eialv oi ABr, AEZ xuxXoi, Taai eialv ai 
ex xfiv xevxptov 8uo 8r] ai BH, Hr 60o xau; E9, 9Z Taai- 
xal yiovia f\ npbc, xai H ywvia xfj 7ip6c; xai O Tar)- pdau; dpa 
f] Br pdoei xfj EZ eaxiv i'ar). xal inel Tar] eaxlv rj Ttpoc; iw 
A ywvia xfj npbz x« A, ojioiov dpa eaxl xo BAr x^rj^ia xG 
EAZ xjir^iaxi- xai eioiv era Tatov euiSeiSv [xfiv Br, EZ]' xa 
8e era Tacov eMeifiv ojioia xjir^axa xuxXtov Taa dXXrjXou; 
eaxiv Taov apa xo BAr xjrrjua xG EAZ. eaxi 8e xal oXoc; 6 
ABr xuxXoc; oXw ifi AEZ xuxXtp Taoc;- Xoinr) apa f) BKr 
Ttepicpepeia xfj EAZ Ttepicpepeia eaxlv i'ar). 

'Ev apa xou; Taou; xuxXoic; ai i'aai yioviai era Taa>v Ttepi- 
cpepeifiv peprjxaaiv, edv xe icpoc; xou; xevxpou; eav xe icpoc; 
xau; uepicpepeiac; fiai peprjxuTai- onep e8ei SeT^ai. 



In equal circles, equal angles stand upon equal cir- 
cumferences whether they are standing at the center or 
at the circumference. 





K L 

Let ABC and DEF be equal circles, and within them 
let BGC and EHF be equal angles at the center, and 
BAG and EDF (equal angles) at the circumference. I 
say that circumference BKC is equal to circumference 
ELF. 

For let BC and EF have been joined. 

And since circles ABC and DEF are equal, their radii 
are equal. So the two (straight-lines) BG, GC (are) equal 
to the two (straight-lines) EH, HF (respectively). And 
the angle at G (is) equal to the angle at H. Thus, the base 
BC is equal to the base EF [Prop. 1.4]. And since the 
angle at A is equal to the (angle) at D, the segment BAC 
is thus similar to the segment EDF [Dei. 3.11]. And 
they are on equal straight-lines [BC and EF] . And simi- 
lar segments of circles on equal straight-lines are equal to 
one another [Prop. 3.24]. Thus, segment BAC is equal to 
(segment) EDF. And the whole circle ABC is also equal 
to the whole circle DEF. Thus, the remaining circum- 
ference BKC is equal to the (remaining) circumference 
ELF. 

Thus, in equal circles, equal angles stand upon equal 
circumferences, whether they are standing at the center 



95 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



'Ev xou; laoic; xuxXoig al era lacov Tiepicpepeuov pepiqxuTai 
ycoviai i'aai dXXf|Xai<; eiaiv, edv xe Tipoc; xou; xevxpou; edv xe 
Tipoc; xau; Tiepicpepeiau; Sai pe(3r)xuTai. 

^ A 





or at the circumference. (Which is) the very thing which 
it was required to show. 

Proposition 27 

In equal circles, angles standing upon equal circum- 
ferences are equal to one another, whether they are 
standing at the center or at the circumference. 

A_ _ n 




'Ev yap looic, xuxXoig xou; ABr, AEZ Ira Tacov Tiepi- 
cpepeifiv iwv Br, EZ Tipoc; y.sv xolc; H, xevxpou; ycoviai 
PePrjxexcoaav ai utio BBT, E9Z, Tipoc; 8e xau; Tiepicpepeiau; 
ai utio BAr, EAZ- Xeyw, oxi f) |iev utio BBT ywvia xfj utio 
E9Z eaxiv larj, f) Se utio BAr xfj utio EAZ eaxiv for). 

Et yap dviaoc; eaxiv f] utio BHr xfj 6ti6 E9Z, [iia auxGv 
^iei£«v eaxiv. eaxw ^eic^cov f] utio BHr, xod auveaxdxco 
Tipoc; xfj BH eu'deia xal iu Tipoc; auxfj ar^eicp x6> H xfj Otio 
E9Z ywvia for) f) Otio BHK- ai Se foai yoviai era lowv 
TiepicpepeiGv pepf|xaaiv, oxav Tipoc; xou; xevxpou; Saiv Tar] 
dpa f] BK TxepLcpepeia xfj EZ Tiepicpepeia. dXXd f] EZ xfj Br 
eaxiv for)- xal f) BK dpa xfj Br eaxiv for] f] eXdxxtov xfj 
\ieiZovv oTiep eaxiv dSuvaxov. oux dpa dviaoc; eaxiv f) utio 
BHr ytovia xfj utio E9Z- for] dpa. xai eaxi xfjc; [lev utio 
BHr fpiaeia f] Tipoc; xw A, xrjc Se utio E9Z fjuiaeia f) Tipoc; 
to A- for] dpa xal f] Tipoc; xw A yovia xfj Tipoc xw A. 

'Ev dpa xou; foou; xuxXou; ai era i'awv Tiepicpepeiov pe- 
Pr]XuTai ywviai i'aai dXXf|Xai<; eiaiv, edv xe Tipoc; xou; xevxpou; 
edv xe Tipoc talc Tiepicpepeiau; CSai pepr]xuiai- oTiep e8ei SeT^ai. 



XT] . 

'Ev xou; foou; xuxXou; ai i'aai eui5elai i'aac; Tiepicpepeiac; 
dcpaipoOai xf]v (iev (iei^ova xfj jiei^ovi xfjv Se eXdxxova xfj 
eXdxxovi. 

'Eaxwaav laoi xuxXoi oi ABr, AEZ, xai ev xou; xuxXou; 
foai eu^elai eaxcoaav ai AB, AE xdc; [ie\ ATB, AZE Tiepi- 
cpepeiac; jiei^ovac; dcpaipouaai xdc; Se AHB, A9E eXdxxovag- 
Xeyoj, oxi f] [iev ArB jiei^wv Tiepicpepeia for] eaxi xfj AZE 
^ei^ovi Tiepicpepeia f) 8e AHB eXdxxtov Tiepicpepeia xfj A9E. 



For let the angles BGC and EHF at the centers G 
and H, and the (angles) BAG and EDF at the circum- 
ferences, stand upon the equal circumferences BC and 
EF, in the equal circles ABC and DEF (respectively) . I 
say that angle BGC is equal to (angle) EHF, and BAC 
is equal to EDF. 

For if BGC is unequal to EHF, one of them is greater. 
Let BGC be greater, and let the (angle) BGK, equal to 
angle EHF, have been constructed on the straight-line 
BG, at the point G on it [Prop. 1.23]. But equal angles 
(in equal circles) stand upon equal circumferences, when 
they are at the centers [Prop. 3.26]. Thus, circumference 
BK (is) equal to circumference EF. But, EF is equal 
to BC. Thus, BK is also equal to BC, the lesser to the 
greater. The very thing is impossible. Thus, angle BGC 
is not unequal to EHF. Thus, (it is) equal. And the 
(angle) at A is half BGC, and the (angle) at D half EHF 
[Prop. 3.20]. Thus, the angle at A (is) also equal to the 
(angle) at D. 

Thus, in equal circles, angles standing upon equal cir- 
cumferences are equal to one another, whether they are 
standing at the center or at the circumference. (Which is) 
the very thing it was required to show. 

Proposition 28 

In equal circles, equal straight-lines cut off equal cir- 
cumferences, the greater (circumference being equal) to 
the greater, and the lesser to the lesser. 

Let ABC and DEF be equal circles, and let AB 
and DE be equal straight-lines in these circles, cutting 
off the greater circumferences ACB and DFE, and the 
lesser (circumferences) AGB and DHE (respectively). I 
say that the greater circumference ACB is equal to the 
greater circumference DFE, and the lesser circumfer- 



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r Z 




H © 

ElXricp-dw ydp T °< xfvxpa iSv xuxXwv xa K, A, xal 
sriie^Eux'dwaav ad AK, KB, AA, AE. 

Kcd inei Taoi xuxXoi eiaiv, Taai rial xal ai sx xfiv 
xsvxpcov 860 8r) ai AK, KB Sua! xdi<; AA, AE Taai eiaiv 
xal |3daic; f) AB pdoei xfj AE Tory ywvia apa f) Otto AKB 
ycovia xfj Otto AAE Tar] eaxiv. ai 8s Taai ycoviai km. Tacov 
Tispicpspeiwv pepf]xaaiv, oxav npoc, xoT? xevxpoi? Saiv Tar) 
apa f] AHB nepicpepsia xfj AGE. saxl 6e xai okoc, 6 ABr 
xuxXog 6Xcp xcp AEZ xuxXcp Taog- xal Xomf] apa f) ArB 
nepicpepsia Xomfj xfj AZE rcspicpspeia i'or) eaxiv. 

'Ev apa xolc; Taoic; xuxXoic; ai Taai euiSriai Taac; m- 
picpspeiac; dcpaipouai xf]v (lev jiei^ova xrj jirii^ovi xf]v 8e 
sXdxxova xrj eXdxxovi- onep e8si SeT^ai. 



ence AGB to (the lesser) DHE. 



C F 




G H 



For let the centers of the circles, K and L, have been 
found [Prop. 3.1], and let AK, KB, DL, and LE have 
been joined. 

And since {ABC and DEF) are equal circles, their 
radii are also equal [Def. 3.1]. So the two (straight- 
lines) AK, KB are equal to the two (straight-lines) DL, 
LE (respectively). And the base AB (is) equal to the 
base DE. Thus, angle AKB is equal to angle DLE 
[Prop. 1.8]. And equal angles stand upon equal circum- 
ferences, when they are at the centers [Prop. 3.26] . Thus, 
circumference AGB (is) equal to DHE. And the whole 
circle ABC is also equal to the whole circle DEF. Thus, 
the remaining circumference ACB is also equal to the 
remaining circumference DFE. 

Thus, in equal circles, equal straight-lines cut off 
equal circumferences, the greater (circumference being 
equal) to the greater, and the lesser to the lesser. (Which 
is) the very thing it was required to show 



'Ev xoi<; Xaoic, xuxXou; xdc; Taac; Ttepicpepriac; Taai cu-deiai 
uitoxrivouaiv. 

A A 




H 



"Eaxcoaav i'aoi xuxXoi oi ABr, AEZ, xai ev auxdic; Taai 
Ttepicpepsiai a7t£iXf]cpiL>Gjaav ai BHr, E0Z, xal ErcriCeuyjiEkoaav 
ai Br, EZ su'deTai- Xeyio, oxi Tar] eaxiv f] Br xfj EZ. 

EiXr'icp'dw yap xa xevxpa xGv xuxXwv, xai eaxw xd K, 
A, xai STieCeux'dwaav ai BK, KT, EA, AZ. 

Kai ETisiTaT) eaxiv f] BHr Ttepicpepeia xfj E9Z Ttepicpepeia, 



Proposition 29 



In equal circles, equal straight-lines subtend equal cir- 
cumferences. 

A D 




G H 



Let ABC and DEF be equal circles, and within them 
let the equal circumferences BGC and EHF have been 
cut off. And let the straight-lines BC and EF have been 
joined. I say that BC is equal to EF. 

For let the centers of the circles have been found 
[Prop. 3.1], and let them be (at) K and L. And let BK, 



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ELEMENTS BOOK 3 



Tor) eoxl xal yoivia f\ uno BKr xrj Otto EAZ. xocl tizzi 1001 
eiolv oi ABr, AEZ xuxXoi, Toai rial xal al ex xc5v xevxptov 
860 8r) ai BK, Kr Bual xalc; EA, AZ I'oai eiolv xal ytovtag 
Ioa<; Tiepiexouaiv pdoi<; apa f) Br pdoei xfj EZ I'or) eoxiv 

'Ev apa xolc Took; xuxXok; xac laac, Ttepicpepeiac; foai 
eu'deTai utcoxcivouoiv ousp e8ei Ssl^ai. 



KC, EL, and LF have been joined. 

And since the circumference BGC is equal to the cir- 
cumference EHF, the angle BKC is also equal to (an- 
gle) ELF [Prop. 3.27]. And since the circles ABC and 
DEF are equal, their radii are also equal [Def. 3.1]. So 
the two (straight-lines) BK, KC are equal to the two 
(straight-lines) EL, LF (respectively) . And they contain 
equal angles. Thus, the base BC is equal to the base EF 
[Prop. 1.4]. 

Thus, in equal circles, equal straight-lines subtend 
equal circumferences. (Which is) the very thing it was 
required to show. 



X'. 

Trjv BodeToav rcepicpepeiav Blxa xejielv. 

A 




a r b 

TCaxco f\ Sodriaa rcepicpepeia f] AAB- 8eT 8rj xrjv AAB 
Ttepicpepeiav 8()(a xefielv. 

'Etie^eux^co f] AB, xal xsxjjirjcrdo hiyv. xaxa xo V, xal 
dno xou T a/jjisbu xrj AB eui&eia rcp6<; op-ddi; f])fdw r) TA, 
xal £7TE^£U)cdtoaav al AA, AB. 

Kal en el lay) eoxlv f] AT xfj TB, xoivr] 8e f] TA, 860 
8/) ai AT, TA 8uai xalg Br, TA I'oai eloiv xal ytovia f] 
utt:6 ArA ywvla xrj (mo BrA far)- op-Qr] yap exaxepa- pdau; 
apa f] AA pdoei xrj AB Xar\ eoxiv. al 8s: I'oai eO'delai laac; 
Ttepicpepeiag dcpaipouai xr]v [ie\ jieii^ova xfj ^el^ovi xrjv 8e 
eXdxxova xrj eXdxxovi- xdi eoxiv exaxepa x«v AA, AB ize- 
pi9£pei£3v eXdxxcov r^ixuxXlou- I'or) apa f] AA Ttepicpepeia xfj 
AB nepicpepela. 

; H apa SoOelaa nepicpepeia 8i/a xex^rjxai xaxa xo A 
ar][ieiov ojiep eBei Ttoirjoai. 



Proposition 30 

To cut a given circumference in half. 



D 




A C B 

Let ADB be the given circumference. So it is required 
to cut circumference ADB in half. 

Let AB have been joined, and let it have been cut in 
half at (point) C [Prop. 1.10]. And let CD have been 
drawn from point C, at right-angles to AB [Prop. 1.11]. 
And let AD, and DB have been joined. 

And since AC is equal to CB, and CD (is) com- 
mon, the two (straight-lines) AC, CD are equal to the 
two (straight-lines) BC, CD (respectively). And angle 
ACD (is) equal to angle BCD. For (they are) each right- 
angles. Thus, the base AD is equal to the base DB 
[Prop. 1.4]. And equal straight-lines cut off equal circum- 
ferences, the greater (circumference being equal) to the 
greater, and the lesser to the lesser [Prop. 1.28]. And the 
circumferences AD and DB are each less than a semi- 
circle. Thus, circumference AD (is) equal to circumfer- 
ence DB. 

Thus, the given circumference has been cut in half at 
point D. (Which is) the very thing it was required to do. 



Xa'. 

'Ev xuxXcp f] ^iev ev iu f)puxuxX[tp yavia op-Qr] eoxiv, f] 8e 
ev xfi ^.eii^ovi x[if]^.axi eXdxxwv op'Sr^, f\ 8e ev xai eXdxxovi 
x^ir^axi jie^wv op'drjc xal era f] ^iev xoO ]j.ei(l,o\)oc, xur]pcxo<; 
Y«v(a (j.elC"v eoxlv op'drjc;, f) 8e xou eXdxxovo<; xur]pcxo<; 
ycovia eXdxxwv op'drjc;. 



Proposition 31 

In a circle, the angle in a semi-circle is a right-angle, 
and that in a greater segment (is) less than a right-angle, 
and that in a lesser segment (is) greater than a right- 
angle. And, further, the angle of a segment greater (than 
a semi-circle) is greater than a right-angle, and the an- 



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Tiaxw xuxXo<; 6 ABrA, 8id|iexpoc; 8e auxoO eaxto f] BT, 
xevxpov 8e to E, xal CTce£eu)edk>aav ai BA, Ar, AA, AE 
Xeyw, oxi f] [iev ev xfi BAr r^ixuxXup yovia f] utco BAr 
opflVj eaxiv, f] 8e ev xw ABr jieiCovi xoO rpixuxXiou x^r^axi 
y«v(a f) utco ABr eXdxxcov eaxiv opiDrjc;, f] 8e ev xw AAr 
eXdxxovi xou f^ixuxXiou x^ir|uaxi ywvia rj Otxo AAr jiei^wv 
eaxiv op-df^. 

'ETieCeu)fdw f) AE, xal Sir^o rj BA era xo Z. 

Kal CTcei iar] eaxiv f) BE xfj EA, Xar\ eaxl xal ywvia f) 
utco ABE xfj utco BAE. TcdXiv, enel lor\ eaxiv f] TE xfj EA, 
iar] eaxl xal f] utco ArE xfj utco TAE- 6Xr] dpa f) Otco BAr 
Suai xdic utio ABr, ArB Xor\ eaxiv. eaxl 8e xal f] Otco ZAr 
exxoc xou ABr xpiyovou 8ual xdig Otco ABr, ArB ycoviau; 
iar)- Iar] dpa xal f] utco BAr y«v[a xfj utco ZAr- op-df] dpa 
exaxepa- f] dpa ev xw BAr f^ixuxXitp ywvia f] utio BAr 
opi^f] eaxiv. 

Kal CTcei xou ABr xpiywvou 8uo ywviai ai utio ABr, 
BAr 8uo op-dSv eXdxxove<; eiaiv, opftr} 8e f] utco BAr, 
eXdxxov dpa op-drjc eaxiv f) utco ABr ywvia- xai eaxiv ev 
x<3 ABr [xeiCovi xou fpixuxXiou xuf^axi. 

Kal CTiel ev xuxXw xexpaTcXeupov eaxi xo ABrA, xCSv Se 
ev xou; xuxXok; xexpaTcXeupwv ai dTcevavxiov ycoviai Suaiv 
opiJau; iaai eiaiv [ai dpa utco ABr, AAr ytoviai 8uaiv opiJau; 
Taa<; eiaiv], xai eaxiv f] utio ABr eXdxxwv opiDrjc' Xoitct) dpa 
f] utco AAr ywvia ^tei^Mv op'drjc; eaxiv xai eaxiv ev iu AAr 
eXdxxovi xou f)(iixuxX[ou x(if|iJiaxi. 

Aeyw, oxi xai f] (jiev xou x^if|^axo<; ywvia f] Tte- 

piexo^evr] utco [xe] xrjc; ABr Ttepicpepeiac xai xfj? Ar eu-deiac; 
[leiCwv eaxiv op'drjc, f] Se xou eXdxxovoc; x^ir](iaxoc; ywvia f\ 
Tcepiexofievr) utco [xe] xrjc; AA[r] Tcepicpepeiac; xal xfj? Ar 
eu-deiac; eXdxxwv eaxiv op'drjc;. xa( eaxiv auxo'dev cpavepov. 
CTcel yap f] utio twv BA, Ar eu-deiwv op'df] eaxiv, f] dpa 
utco xrjc; ABr Tcepicpepeiac; xal xrjc; Ar eui9eia<; Ttepiexo^tevr] 
^ei£tov eaxiv op'drjc;. TidXiv, CTcel f] utco xwv Ar, AZ euTJeiSv 
op'dr] eaxiv, f] dpa utco xrjc; IA eui!)eia<; xai xrjc; AA[r] rcepi- 



gle of a segment less (than a semi-circle) is less than a 
right-angle. 




Let ABCD be a circle, and let BC be its diameter, and 
E its center. And let BA, AC, AD, and DC have been 
joined. I say that the angle BAG in the semi-circle BAC 
is a right-angle, and the angle ABC in the segment ABC, 
(which is) greater than a semi-circle, is less than a right- 
angle, and the angle ADC in the segment ADC, (which 
is) less than a semi-circle, is greater than a right-angle. 

Let AE have been joined, and let BA have been 
drawn through to F. 

And since BE is equal to EA, angle ABE is also 
equal to BAE [Prop. 1.5]. Again, since CE is equal to 
EA, ACE is also equal to CAE [Prop. 1.5]. Thus, the 
whole (angle) BAC is equal to the two (angles) ABC 
and ACB. And FAC, (which is) external to triangle 
ABC, is also equal to the two angles ABC and ACB 
[Prop. 1.32]. Thus, angle BAC (is) also equal to FAC. 
Thus, (they are) each right-angles. [Def. 1.10]. Thus, the 
angle BAC in the semi-circle BAC is a right-angle. 

And since the two angles ABC and BAC of trian- 
gle ABC are less than two right-angles [Prop. 1.17], and 
BAC is a right-angle, angle ABC is thus less than a right- 
angle. And it is in segment ABC, (which is) greater than 
a semi-circle. 

And since ABCD is a quadrilateral within a circle, 
and for quadrilaterals within circles the (sum of the) op- 
posite angles is equal to two right-angles [Prop. 3.22] 
[angles ABC and ADC are thus equal to two right- 
angles], and (angle) ABC is less than a right-angle. The 
remaining angle ADC is thus greater than a right-angle. 
And it is in segment ADC, (which is) less than a semi- 
circle. 

I also say that the angle of the greater segment, 
(namely) that contained by the circumference ABC and 
the straight-line AC, is greater than a right-angle. And 
the angle of the lesser segment, (namely) that contained 



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cpepdac; 7tepie)(o^ev7) eXdxxcov eoxlv 6piDfj<;. 

'Ev xuxXtp apa f\ [iev s\ xG fjjiixuxXicp yiovia 6p^>r) eaxiv, 
f) 8s ev xG ^te[£ovi xuf|pcxi eXdxxwv opiS/jt;, f) 8e ev xG 
eXdxxovi [x^tr^iaxi] [icii^wv op-dfjc xal era f] (iev xou ^.el^ovoc; 
x^.r][iaxo(; [ycovla] (ici^ov [eaxlv] op-drjc;, f) 8s xou eXdxxovoc 
x^r^iaxoc; [yiovia] eXdxxtov op-dfje onep e8ei Set^ai. 



X(3'. 

'Eav xuxXou scpdnxrjxai xu; euiDela, duo 8s xfjc; dcpfji; ei<; 
xov xuxXov 8ia)cdr) xi<; eu-dela xe^ivouaa xov xuxXov, a<; 
Tioiel yoviac Txpog xfj e(pan;xo^evr], faai eaovxai xalc; ev xou; 
evaXXdc; xou xuxXou x^ur^aai ywvlait;. 




KuxXou yap xou ABrA ecpanxeo'dw xu; eu-dela f\ EZ 
xaxd xo B ar^elov, xal duo xou B ar^dou Sir^^co T ^ 
cu-dela zlc, xov ABrA xuxXov xe^ivouaa auxov rj BA. Xeyto, 
6xi a<; TioieT ywv[a<; f] BA ^exa xrj<; EZ ecpaTtxo^evr]<;, l'aa<; 
saovxai xdu; ev xolg evaXXdJ; xjiruiaoi xou xuxXou yoviau;, 
xouxeaxiv, oxi f) \ie\ Otto ZBA ycovia Xar] eaxl xrj sv xG BAA 
x|ir]jiaxi ouviaxajisv/) ywvia, f] 8e utto EBA y«v[a Tar) eaxl 
xfj ev xG ArB xjj.iq[iaxi auviaxa(j.evr) ycovia. 

"Hy^co yap duo xou B xfj EZ Ttpoc; opiSat; f] BA, xal 
eiXiQcp'dw era xfj<; BA 7iepicpepe(a<; xu)(6v ar)[ieTov xo T, xal 
sne^sux'dwaav ai AA, Ar, TB. 

Kal enel xuxXou xou ABrA ecpdnxexai xu; eMela f) EZ 



by the circumference AD[C] and the straight-line AC, is 
less than a right-angle. And this is immediately apparent. 
For since the (angle contained by) the two straight-lines 
BA and AC is a right-angle, the (angle) contained by 
the circumference ABC and the straight-line AC is thus 
greater than a right-angle. Again, since the (angle con- 
tained by) the straight-lines AC and AF is a right-angle, 
the (angle) contained by the circumference AZ)[C] and 
the straight-line CA is thus less than a right-angle. 

Thus, in a circle, the angle in a semi-circle is a right- 
angle, and that in a greater segment (is) less than a 
right-angle, and that in a lesser [segment] (is) greater 
than a right-angle. And, further, the [angle] of a seg- 
ment greater (than a semi-circle) [is] greater than a right- 
angle, and the [angle] of a segment less (than a semi- 
circle) is less than a right-angle. (Which is) the very thing 
it was required to show. 

Proposition 32 

If some straight-line touches a circle, and some 
(other) straight-line is drawn across, from the point of 
contact into the circle, cutting the circle (in two), then 
those angles the (straight-line) makes with the tangent 
will be equal to the angles in the alternate segments of 
the circle. 

A 




For let some straight-line EF touch the circle ABCD 
at the point B, and let some (other) straight-line BD 
have been drawn from point B into the circle ABCD, 
cutting it (in two) . I say that the angles BD makes with 
the tangent EF will be equal to the angles in the alter- 
nate segments of the circle. That is to say, that angle 
FBD is equal to the angle constructed in segment BAD, 
and angle EBD is equal to the angle constructed in seg- 
ment DCB. 

For let BA have been drawn from B, at right-angles 
to EF [Prop. 1.11]. And let the point C have been taken 
at random on the circumference BD. And let AD, DC, 



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xaxd to B, xal dmo xfjc; acpfjc; rjxxai xfj ecpauxo^iivr] Ttpoc; 
opMc; f] BA, era xfjc BA apa xo xevxpov eaxl xou ABrA 
xuxXou. f] BA apa Bid^texoc; eaxi xou ABrA xuxXou- f] apa 
uno AAB ywvia ev f)|iixuxX[fc> ouaa op'dr] eaxiv. Xoiraxi apa 
ai bnb BAA, ABA ^tia op'dfj i'aai eiaiv. eoxl Be xal f] Otto 
ABZ op'Ory fj apa (mo ABZ iar) eaxi xdic; Otto BAA, ABA. 
xoivfj dcprprp'dco f) Otto ABA- XoiTtf] dpa f) 0ti6 ABZ ywvia 
Xat] eaxl xfj ev xG evaXXdJ; xjiiq^axi xou xuxXou ycovia xfj 
Otto BAA. xal end ev xuxXip xexpaTiXeupov eaxi xo ABrA, 
ai dnevavxiov aOxou ywviai Sualv op^aTc; i'aai eiaiv. eiol Be 
xal ai Otto ABZ, ABE Bualv opiate; i'aai- ai apa Otto ABZ, 
ABE xdic; Otto BAA, BEA iaai eiaiv, Sv f] Otto BAA xfj Otto 
ABZ eSei/i}/) far) - Xomf] dpa f) utio ABE xfj ev xG evaXXdi; 
xou xuxXou xjir|(jiaxi xG ArB xfj Otto ArB yovia eaxiv Tar). 

°Eav dpa xuxXou ecpaTtxrjxai xic; eu-dela, diro Be xfjc; acpfjc; 
eic; xov xuxXov 8ia)c8f) xic; eui5ela xejivouaa xov xuxXov, ac; 
TioieT ycoviac; Tip6<; xfj ecpauxo^ievr], iaai eaovxai xdic; ev xolc; 
evaXXac; xou xuxXou xjnqjiaai ywviaic;- OTtep eBei Bel^ai. 



and CB have been joined. 

And since some straight-line EF touches the circle 
ABCD at point B, and BA has been drawn from the 
point of contact, at right-angles to the tangent, the center 
of circle ABCD is thus on BA [Prop. 3.19]. Thus, BA 
is a diameter of circle ABCD. Thus, angle ADB, being 
in a semi-circle, is a right-angle [Prop. 3.31]. Thus, the 
remaining angles (of triangle ADB) BAD and ABD are 
equal to one right-angle [Prop. 1.32]. And ABF is also a 
right-angle. Thus, ABF is equal to BAD and ABD. Let 
ABD have been subtracted from both. Thus, the remain- 
ing angle DBF is equal to the angle BAD in the alternate 
segment of the circle. And since ABCD is a quadrilateral 
in a circle, (the sum of) its opposite angles is equal to 
two right-angles [Prop. 3.22]. And DBF and DBF is 
also equal to two right-angles [Prop. 1.13]. Thus, DBF 
and DBF is equal to BAD and BCD, of which BAD 
was shown (to be) equal to DBF. Thus, the remaining 
(angle) DBF is equal to the angle DCB in the alternate 
segment DCB of the circle. 

Thus, if some straight-line touches a circle, and some 
(other) straight-line is drawn across, from the point of 
contact into the circle, cutting the circle (in two), then 
those angles the (straight-line) makes with the tangent 
will be equal to the angles in the alternate segments of 
the circle. (Which is) the very thing it was required to 
show. 



Ay'. 

'Era xfjc; Bo-Mo/]? cu^eiac; ypd^ai xufjpt xuxXou Bexo^ie- 
vov yoviav ia/]v xrj Bo'deiar] ywvia etrduypdu^w. 




E b E 



"Eaxw f) Bo'Oelaa euiJeTa f] AB, f) Be BoiJeTaa yovia 
eO-duypajjijioc; f] Ttpoc; xG E Bel Br) era xfjc; Bo-deiarjc; eO-deiac; 
xfjc AB ypd(j;ai xjifj^ia xuxXou Be/ojjievov ywviav larjv xrj 
Ttpoc; xG r. 

H Bf) Ttpoc; xG r [ytovia] rjxoi 6c;e1d eaxiv f\ op'df) fj 
djjipXeTa- eaxw Ttpoxepov 6i;eTa, xal Gc; era xfjc; TtpGxrjc; xa- 
xaypacpfjc auveaxdxw Ttpoc; xfj AB eO'deia xal xG A ar)[is'\.(x> 
xfj Ttpoc; xG r ywvia iar) f) Otto BAA' 6<;eTa dpa eaxl xal f) 
utio BAA. f]xi3« xfj AA Ttpoc; opftac, f] AE, xal xexjif)ai9« 
f] AB 8[)(a xaxd xo Z, xal fjX'&co dno xou Z arj^ebu xfj AB 



Proposition 33 

To draw a segment of a circle, accepting an angle 
equal to a given rectilinear angle, on a given straight-line. 




E B E 



Let AB be the given straight-line, and C the given 
rectilinear angle. So it is required to draw a segment 
of a circle, accepting an angle equal to C, on the given 
straight-line AB. 

So the [angle] C is surely either acute, a right-angle, 
or obtuse. First of all, let it be acute. And, as in the first 
diagram (from the left), let (angle) BAD, equal to angle 
C, have been constructed on the straight-line AB, at the 
point A (on it) [Prop. 1.23]. Thus, BAD is also acute. Let 
AE have been drawn, at right-angles to DA [Prop. 1.11]. 



101 



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ELEMENTS BOOK 3 



Kpbc, op-ddc f] ZH, xal STte^eux'&o f] HB. 

Kal enel far) eaxiv f) AZ xfj ZB, xoivf) 8e f) ZH, 860 8f) 
ai AZ, ZH Suo xdic BZ, ZH faai eiaiv xal yovia f) utio 
AZH [yov[a] xfj utio BZH far)- pdaic dpa f) AH pdaei xfj 
BH lor] eaxiv. 6 dpa xevxpo ^.ev iw H 8iaaxr)[iaxi Se iu 
HA xuxXoc ypacpouevoc fjcei xal 8id xou B. yeypdqydo x °d 
eaxo 6 ABE, xal CTieCeux'do ^ EB. excel ouv dm' axpac xfjc 
AE Sia^iexpou duo xou A xrj AE xcpoc op'ddc eaxiv f) AA, 
f] A A apa ecpdicxexai xou ABE xuxXou- excel ouv xuxXou 
xou ABE ecpdxcxexai xu; eu'dela f) AA, xal duo xfjc xaxd xo 
A dcpfjc tic, xov ABE xuxXov Sifjxxai tic, cu^ela f) AB, f] 
apa utio AAB yovia for) eaxl xfj ev tw evaXXdc' xou xuxXou 
x^ir)[iaxi yovia xfj utio AEB. dXX' r\ utio AAB xfj Tipoc xo T 
eaxiv for)- xal f) Tipoc xo T apa yovia for] eaxl xfj utio AEB. 

Tkl xfjc So-deiarjc apa eui!)e(a<; xfjc AB x^if)[ia xuxXou 
yeypaKxai xo AEB Sexojievov yoviav xf]v utio AEB I'arjv 
xrj Bo'deiorj xfj Ttpoc xo T. 

AXXa 8f] op'df] eaxo f) Ttpoc xo E xal Seov TidXiv eaxo 
excl xfjc AB ypdtjjai xjifjpt xuxXou Se)(6fievov yoviav Tarjv xfj 
xcpoc xo T op'drj [yovia]. auveaxdxo [xcdtXiv] xrj Tcpoc xo T 
dp'&r] yovia I'ar) f] utio BAA, 6<; e^ei excl xfjc 8euxepac xaxa- 
ypacpfjc, xal xex^.rfo'do f) AB 8[)(a xaxd xo Z, xal xevxpo xo 
Z, 8iaaxf]^axi Se oxcoxepo xov ZA, ZB, xuxXoc yeypdcp-do 
6 AEB. 

'Ecpduxexai apa f] AA eu'dela xou ABE xuxXou Bid xo 
op'drjv eTvai xf]v Tcpoc xo A yoviav. xal for) eaxiv f] utio 
BAA yovia xrj ev xo AEB x^r^axi- op'df] yap tod auxrj ev 
f]^ixuxXio ouaa. dXXd xal f] utio BAA xfj Tipoc xo T far) 
eaxiv. xal f] ev to AEB apa far) lax! xfj Tipoc xo F. 

TcypaTixai apa TidXiv era xfjc AB x^tfjua xuxXou xo AEB 
8e)(6^evov yoviav larjv xfj Tipoc xo T. 

AXXa 8f) f) Tip6<; xo T d^pXsTa eaxo- xal auvsaxdxo 
auxfj larj Tipoc xfj AB euiJeia xal xo A ar)[ie(o f) utio BAA, 
6c ex £l T ^ tpttTiZ xaxaypacpfjc, xal xfj AA Tipoc op'&a.z 
fjx'do f) AE, xal xex^rjado TidXiv f) AB 8i^a xaxd xo Z, xal 
xfj AB Tip6<; op-ddc fjx'do f) ZH, xal £TieCeu)fdo rj HB. 

Kal ETtd TidXiv far) saxlv f) AZ xfj ZB, xal xoivf) f) ZH, 
8uo 8fj ai AZ, ZH Suo xdl<; BZ, ZH faai eiaiv xal yovia f) 
utio AZH yovla xfj utio BZH far)- pdaig apa f) AH pdaei 
xfj BH far] eaxiv 6 apa xevxpo ^xsv xo H 8iaaxr)(jiaxi 8s xo 
HA xuxXo<; ypacp6^ie:vo<; fj^Ei xal 8id xou B. ep^sa'do 6<; 6 
AEB. xal etieI xfj AE Bia^iexpo dii' dxpac Tipoc op-dac. eaxiv 
f) AA, f) AA apa ecpdiixexai xou AEB xuxXou. xal diio xfj<; 
xaxd xo A eTiacpfjc Sifjxxai f) AB- f) apa utio BAA yovia 
far) eaxl xfj ev xo evaXXd^ xou xuxXou x^irj^iaxi xo A0B 
auviaxa^ifvr) yovia. dXX' f) utio BAA yovia xfj Tip6<; xo T 
far) eaxiv. xal f) ev xo A0B apa x^rpaxi yovia far) eaxl xfj 
Tip6<; xo r. 

'EtiI xfjc apa ScMar^ euiJeiac xfjc AB yeypaiixai x^fjjia 
xuxXou xo A0B 8exo^ievov yoviav far)v xfj Tipoc xo T- oTiep 
e8ei Tioifjaai. 



And let AB have been cut in half at F [Prop. 1.10]. And 
let FG have been drawn from point F, at right-angles to 
AB [Prop. 1.11]. And let GB have been joined. 

And since AF is equal to FB, and FG (is) common, 
the two (straight-lines) AF, FG are equal to the two 
(straight-lines) BF, FG (respectively). And angle AFG 
(is) equal to [angle] BFG. Thus, the base AG is equal to 
the base BG [Prop. 1.4]. Thus, the circle drawn with 
center G, and radius GA, will also go through B (as 
well as A). Let it have been drawn, and let it be (de- 
noted) ABE. And let EB have been joined. Therefore, 
since AD is at the extremity of diameter AE, (namely, 
point) A, at right-angles to AE, the (straight-line) AD 
thus touches the circle ABE [Prop. 3.16 corr.]. There- 
fore, since some straight-line AD touches the circle ABE, 
and some (other) straight-line AB has been drawn across 
from the point of contact A into circle ABE, angle DAB 
is thus equal to the angle AEB in the alternate segment 
of the circle [Prop. 3.32]. But, DAB is equal to C. Thus, 
angle C is also equal to AEB. 

Thus, a segment AEB of a circle, accepting the angle 
AEB (which is) equal to the given (angle) C, has been 
drawn on the given straight-line AB. 

And so let C be a right-angle. And let it again be 
necessary to draw a segment of a circle on AB, accepting 
an angle equal to the right- [angle] C. Let the (angle) 
BAD [again] have been constructed, equal to the right- 
angle C [Prop. 1.23], as in the second diagram (from the 
left). And let AB have been cut in half at F [Prop. 1.10]. 
And let the circle AEB have been drawn with center F, 
and radius either FA or FB. 

Thus, the straight-line AD touches the circle ABE, on 
account of the angle at A being a right-angle [Prop. 3.16 
corr.]. And angle BAD is equal to the angle in segment 
AEB. For (the latter angle), being in a semi-circle, is also 
a right-angle [Prop. 3.31]. But, BAD is also equal to C. 
Thus, the (angle) in (segment) AEB is also equal to C. 

Thus, a segment AEB of a circle, accepting an angle 
equal to C, has again been drawn on AB. 

And so let (angle) C be obtuse. And let (angle) BAD, 
equal to (C), have been constructed on the straight-line 
AB, at the point A (on it) [Prop. 1.23], as in the third 
diagram (from the left) . And let AE have been drawn, at 
right-angles to AD [Prop. 1.11]. And let AB have again 
been cut in half at F [Prop. 1.10] . And let FG have been 
drawn, at right-angles to AB [Prop. 1.10]. And let GB 
have been joined. 

And again, since AF is equal to FB, and FG (is) 
common, the two (straight-lines) AF, FG are equal to 
the two (straight-lines) BF, FG (respectively). And an- 
gle AFG (is) equal to angle BFG. Thus, the base AG is 



102 



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ELEMENTS BOOK 3 




"Eaxw 6 Bo'delc; xuxXoc; 6 ABr, f] Be Bo'deTaa ywvia 
eu-duYpa^cx; f] npoc, xG A- Bei 8f| &7t6 xou ABr xuxXou 
T^fjfia dcpeXeTv Sexo^Evov yaviav I'arjv xfj Bo'defay] ywv'a 
£ui}uYpd(ijj.« xfj Tipog xfi A. 

"Hx^w toO ABr ecpaTTTojisvr) f] EZ xaxd xo B ar^aov, 
xal guvsgx&xo npoc xfj ZB eu-dda xal xG 7ipo<; auxfj ar^do 
xG B xfj Tipoc; xG A ywvia Tar] f] 0n:6 ZBI\ 

'End ouv xuxXou xou ABr dpdnxexai xic su^da f) EZ, 
xal duo xrjc; xaxd xo B dxacprjc Birjxxai f) Br, f] uizb ZBT apa 
yovia Tar] eaxi xfj sv xG BAr evaXXd?; x[if]^.axi auviaxa^iivr] 
yovia. dXX' f) Otto ZBr xrj npoc xG A eaxiv Xor\ % xal f] ev 
xG BAr apa x(if|jj.axi lot) eoxl xfj 7tp6<; xG A [ywvia] . 

Ako xou So-devxot; apa xuxXou xou ABr x^ifj^a dcpfjprjxai 
xo BAr 8e)(6^£vov ywviav I'arjv xfj BoiJeior] ywvia su'duYpd^i- 
[ia> xfj npoc; iu A- oTiep eSei noirjaai. 



t Presumably, by finding the center of ABC [Prop. 3.1], drawing a 



equal to the base BG [Prop. 1.4]. Thus, a circle of center 
G, and radius GA, being drawn, will also go through B 
(as well as A) . Let it go like AEB (in the third diagram 
from the left). And since AD is at right-angles to the di- 
ameter AE, at its extremity, AD thus touches circle AEB 
[Prop. 3.16 corr.]. And AB has been drawn across (the 
circle) from the point of contact A. Thus, angle BAD is 
equal to the angle constructed in the alternate segment 
AEB of the circle [Prop. 3.32]. But, angle BAD is equal 
to C. Thus, the angle in segment AHB is also equal to 
C. 

Thus, a segment AHB of a circle, accepting an angle 
equal to C, has been drawn on the given straight-line AB. 
(Which is) the very thing it was required to do. 

Proposition 34 

To cut off a segment, accepting an angle equal to a 
given rectilinear angle, from a given circle. 




Let ABC be the given circle, and D the given rectilin- 
ear angle. So it is required to cut off a segment, accepting 
an angle equal to the given rectilinear angle D, from the 
given circle ABC. 

Let EF have been drawn touching ABC at point BJ 
And let (angle) FBC, equal to angle D, have been con- 
structed on the straight-line FB, at the point B on it 
[Prop. 1.23]. 

Therefore, since some straight-line EF touches the 
circle ABC, and BC has been drawn across (the circle) 
from the point of contact B, angle FBC is thus equal 
to the angle constructed in the alternate segment BAG 
[Prop. 1.32]. But, FBC is equal to D. Thus, the (angle) 
in the segment BAC is also equal to [angle] D. 

Thus, the segment BAC, accepting an angle equal to 
the given rectilinear angle D, has been cut off from the 
given circle ABC. (Which is) the very thing it was re- 
quired to do. 

ht-line between the center and point B, and then drawing EF through 



103 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



point B, at right-angles to the aforementioned straight-line [Prop. 1.11]. 

Xe'. 



'Eav sv xuxXcp 860 su-dslai x£^iv«aiv dXXfjXac;, to utco 
t£>v Tfjc; (iia<; T^irpaTCOv Tcspisxojisvov op'doytoviov i'aov sail 
to utco t65v Tfjc; STspac; T^r)[idTWv Tispis/ojievcp op-doytoviw. 




'Ev yap xuxXw to ABEA 860 sMsTai ai Ar, BA 
TS^vsTtoaav dXXfjXac; xaxd to E ar][Mov Xeyco, oxi to utco 
twv AE, Er Tcspisxojisvov 6piL>oy«viov Iaov saTi tQ utco 
twv AE, EB Tcspisxo|iEvw op-doycovio. 

El jjisv ouv ai Ar, BA 81a tou xsvTpou siaiv wots to E 
xevTpov slvai tou ABEA xuxXou, cpavspov, oti iacov ouaGv 
iSv AE, Er, AE, EB xod to utco tGv AE, Er Tcspisxojisvov 
op^oycoviov Iaov laxl tw utco xwv AE, EB Tcspisxo^svw 
op^oyMvicp. 

Mr] saToaav 8f) ai Ar, AB 81a tou xsvTpou, xai 
eiXrjcp'dw to xsvTpov tou ABrA, xai saTM to Z, xai dico 
tou Z era to«; Ar, AB suifteiac; xordeToi fjx'dwcrav ai ZH, 
Z6, xai STieCeuxtiwaav ai ZB, Zr, ZE. 

Kai stcsi suiMd tic; 81a tou xsvTpou f] HZ su-fMav Tiva 
\±r\ Sid tou xevTpou ttjv Ar Tcpoc; opMc; ts^ivsi, xai Bi^a 
ai)Tf|V ts[ivsi- i'ar) apa f] AH Tfj Hr. stcei ouv su'dsTa f) Ar 
T£T^r)Tai sic; [isv laa xaTa to H, sic 8s aviaa xaTa to E, to 
apa utco itov AE, Er Tcspisxo^isvov opiDoywviov ^.STa tou 
and Tfj<; EH TSTpaywvou 'iaov taxi tw duo Tfjc; Hr- [xoivov] 
Tcpoaxsia'dw to dico xrj<; HZ- to apa utco tGv AE, Er ^STa 
twv dico twv HE, HZ i'aov scrci toTc; dico tGv TH, HZ. dXXa 
toTc [Jtsv duo twv EH, HZ Iaov saTi to duo Tfjc; ZE, tou; 
8s duo twv TH, HZ Iaov scrci to duo Tfjc; ZE to apa utco 
twv AE, Er ^.STa tou aKO Tfjc; ZE i'aov scrci tw dico Tfjc; 
Zr. i'ar) 8s f] ZT Tfj ZB- to apa utco twv AE, Er jxexa tou 
dico Tfjc; EZ Iaov saTi tw duo Tfjc; ZB. 81a Ta aika Sf] xai 
to utco tov AE, EB ^.STa tou dico Tfjc; ZE iaov saTi to dico 
Tfjc; ZB. sBsix'dr) 8s xai to utco tGv AE, Er ^iSTa tou duo 
Tfjc; ZE i'aov to dico Tfjc; ZB- to apa utco icov AE, Er ^iSTa 
tou aTco Tfjc; ZE 'iaov sot! tu utco twv AE, EB ^tSTa tou 
dico Tfjc; ZE. xoivov dcpfjpf]a'do to dico Tfjc; ZE- Xoitcov apa 
to utco twv AE, Er Tcspisxo^tsvov op-doycoviov iaov saTi to 
utco tCSv AE, EB Tcspisxo^svw op^oywviw. 

'Eav apa sv xuxXcp su-dslai 860 TS^tvoaiv dXXfjXac;, to 
utco tuv Tfjc fjudg T^trpaTOv Tcspisxo^xsvov opTJoywviov iaov 



Proposition 35 

If two straight-lines in a circle cut one another then 
the rectangle contained by the pieces of one is equal to 
the rectangle contained by the pieces of the other. 




For let the two straight-lines AC and BD, in the circle 
ABCD, cut one another at point E. I say that the rect- 
angle contained by AE and EC is equal to the rectangle 
contained by DE and EB. 

In fact, if AC and BD are through the center (as in 
the first diagram from the left), so that E is the center of 
circle ABCD, then (it is) clear that, AE, EC, DE, and 
EB being equal, the rectangle contained by AE and EC 
is also equal to the rectangle contained by DE and EB. 

So let AC and DB not be though the center (as in 
the second diagram from the left), and let the center of 
ABCD have been found [Prop. 3.1], and let it be (at) F. 
And let FG and FH have been drawn from F, perpen- 
dicular to the straight-lines AC and DB (respectively) 
[Prop. 1.12]. And let FB, FC, and F E have been joined. 

And since some straight-line, GF, through the center, 
cuts at right-angles some (other) straight-line, AC, not 
through the center, then it also cuts it in half [Prop. 3.3]. 
Thus, AG (is) equal to GC. Therefore, since the straight- 
line AC is cut equally at G, and unequally at E, the 
rectangle contained by AE and EC plus the square on 
EG is thus equal to the (square) on GC [Prop. 2.5]. Let 
the (square) on GF have been added [to both]. Thus, 
the (rectangle contained) by AE and EC plus the (sum 
of the squares) on GE and GF is equal to the (sum of 
the squares) on CG and GF. But, the (square) on FE 
is equal to the (sum of the squares) on EG and GF 
[Prop. 1.47], and the (square) on FC is equal to the (sum 
of the squares) on CG and GF [Prop. 1.47]. Thus, the 
(rectangle contained) by AE and EC plus the (square) 
on FE is equal to the (square) on FC. And FC (is) 
equal to FB. Thus, the (rectangle contained) by AE 
and EC plus the (square) on FE is equal to the (square) 
on FB. So, for the same (reasons), the (rectangle con- 
tained) by DE and EB plus the (square) on FE is equal 



104 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



kail xo Otto x«v xfjc sxspag x|ir]jidx«v 7iepiex°^ vt P 0?$°- 
ycovicp- oizep eSei Sei^ai. 



Xt'. 

'Edv xuxXou Xr)cpa[}fj xi ar^elov exioc,, xal dm' auxoO 
Ttpoc; xov xuxXov TtpoaTUTtxoai 860 s&dsTou, xal f] ^tev auxov 
xe^vr] tov xuxXov, f\ 8s Ecpdmxrjxai, saxai to Otto oXiqc; xrj<; 
T£[ivo6ar)c; xal iff exxog aTraXa|jipavo|jievr]c; jisxa^u xou is 
a/]ue(ou xal xfjt: xupxrjc; Ttepicpepslat; I'aov to duo iff ccpa- 
iXTO^isvrjc; xexpayovcp. 




A 



KuxXou yap toO ABr EiXr]cp'dco ti arj^eTov exxoc; to A, 
xal and xou A Ttpoc; tov ABr xuxXov TtpocnuTtxexcoaav 860 
eO'delai ai Ar[A], AB- xal f\ [isv ATA xeuvexo tov ABr 
xuxXov, f] 8s BA scpaTtxecrdw Xeyw, oil to utio xov AA, 
Ar ispisxo[is\iov opiJoyoviov laov kail to aTto xfj? AB 
xexpayovo. 

TL dpa [A] FA fjxoi 81a tou xevxpou eaxlv fj ou. saxo 
TipoTspov Sid tou xevxpou, xal eaxo to Z xevxpov tou ABr 
xuxXou, xal eTC^euy^co f) ZB- 6pi9r] dpa eaxlv f] utio ZBA. 
xal ETtel euiJJsTa #j Ar Stya xex^rjxai xaxd to Z, Ttpoaxeixai 
6e aOxfj T) TA, to dpa utio xov AA, Ar \xsia tou duo iff 
Zr laov sail to dno xfjt; ZA. Tar) Ss f] Zr xfj ZB- xo dpa 
utio xov A A, Ar [isxa xou dito xfjc ZB laov eoxl xo aTto 
xr]<; ZA. xo 6e aTto xrjg ZA Xaa sail xd dito xov ZB, BA- 
xo dpa utio xov AA, Ar [ieia xou dno xfjc ZB Taov eoxl 
xolc; aTto xov ZB, BA. xoivov acpflprjado xo aTto xfjt; ZB- 
XoiTtov dpa xo utio xov AA, Ar ioov tail xo aTto xrjc; AB 



to the (square) on FB. And the (rectangle contained) 
by AE and EC plus the (square) on FE was also shown 
(to be) equal to the (square) on FB. Thus, the (rect- 
angle contained) by AE and EC plus the (square) on 
FE is equal to the (rectangle contained) by DE and EB 
plus the (square) on FE. Let the (square) on FE have 
been taken from both. Thus, the remaining rectangle con- 
tained by AE and EC is equal to the rectangle contained 
by DE and EB. 

Thus, if two straight-lines in a circle cut one another 
then the rectangle contained by the pieces of one is equal 
to the rectangle contained by the pieces of the other. 
(Which is) the very thing it was required to show. 

Proposition 36 

If some point is taken outside a circle, and two 
straight-lines radiate from it towards the circle, and (one) 
of them cuts the circle, and the (other) touches (it), then 
the (rectangle contained) by the whole (straight-line) 
cutting (the circle), and the (part of it) cut off outside 
(the circle), between the point and the convex circumfer- 
ence, will be equal to the square on the tangent (line) . 




D 



For let some point D have been taken outside circle 
ABC, and let two straight-lines, DC[A] and DB, radi- 
ate from D towards circle ABC. And let DC A cut circle 
ABC, and let BD touch (it). I say that the rectangle 
contained by AD and DC is equal to the square on DB. 

[D]CA is surely either through the center, or not. Let 
it first of all be through the center, and let F be the cen- 
ter of circle ABC, and let FB have been joined. Thus, 
(angle) FED is a right-angle [Prop. 3.18]. And since 
straight-line AC is cut in half at F, let CD have been 
added to it. Thus, the (rectangle contained) by AD and 
DC plus the (square) on FC is equal to the (square) on 
FD [Prop. 2.6]. And FC (is) equal to FB. Thus, the 
(rectangle contained) by AD and DC plus the (square) 
on FB is equal to the (square) on FD. And the (square) 
on FD is equal to the (sum of the squares) on FB and 
BD [Prop. 1.47]. Thus, the (rectangle contained) by AD 



105 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



ecpan:xojievr)c. 

AXXd 8f) f\ ATA [L7] eaxw Bid xou xevxpou xou ABr 
xuxXou, xal eiXfjcpi^Gj to xevxpov to E, xal duo xou E eicl 
xf)v Ar xd-dexoc ryyj}^ f] EZ, xal eTte£eu)edioaav ai EB, Er, 
EA- op'&r] apa eaxlv f) Otto EBA. xal enel eu-deld xic 8id xou 
xevxpou f) EZ eu'deTdv xiva [if] Sid xou xevxpou xfjv Ar npoc 
opiDdt; xejivei, xal 8[)(a auxf]v xe^tver f\ AZ dpa xfj Zr eaxiv 
tar), xal end eO'deTa f] Ar xex|i/]xai 8[)(a xaxd xo Z arj^slov, 
icpoaxeixai 8e auxfj f] TA, xo dpa Otto ifiv AA, Ar ^texd xou 
duo xfjc Zr ioov eaxl x£i dno xfjc ZA. xoivov TipoaxeLai&M 
xo diio xfjc ZE- xo dpa Otto xov AA, Ar ^.exd xwv duo xwv 
TZ, ZE laov eaxl xolc duo xwv ZA, ZE. xolc 8s duo xfiv 
rZ, ZE laov eaxl xo duo xfjc Er- 6pi}f] yap [eaxiv] f\ Otto 
EZr [ycovia]- xolc 8s duo xov AZ, ZE I'aov eaxl xo duo xfjc 
EA- xo apa Otto x«v AA, Ar jiexd xou dno xfjc Er I'aov 
eaxl xw aTio xfjc; EA. for) 8e f) Er xfj EB' xo dpa Otto x65v 
AA, Ar jiexa xou diro xfjc EB laov eaxl xo aTro xfjc EA. 
tu 8s dTto xfjc EA laa eaxl xd duo xGv EB, BA- op'df] yap 
f\ Otto EBA ytovia- xo apa Otto xwv AA, Ar ^.exd xou dTto 
xfjc EB laov eaxl xolc duo xwv EB, BA. xoivov dcpypfja'do 
xo ctnb xfjc EB- Xoittov dpa xo Otto xwv AA, Ar laov eaxl 
xo duo xfjc AB. 

Edv dpa xuxXou Xrjcp'dfj xi or\\±eiov exxoc, xal dm' auxou 
Tipoc xov xuxXov TipoaTiiTixwai Suo eu'delai, xal f] ^tev auxcov 
xe^ivr] xov xuxXov, f\ Be ecpdux/jxai, eaxai xo Otto oX/]c xfjc 
xe^vouarjc xal xfjc exxoc diioXa^pavo^evr]c [icxacu xou xe 
ar)[ie(ou xal xfjc xupxfjc icepicpepeiac laov xw duo xfjc ecpa- 
Ttxo^ievr)c xexpaywvy oicep eSei Belial. 



AC- 

°Edv xuxXou Xr)(pi9fi xi a/jjielov exxoc, duo 8e xou 
ar][ieiou Tcpoc xov xuxXov upoaTciTcxwai 860 euiMai, xal 
f) y.ev auxwv xejivr) xov xuxXov, f] 8e TipoaTUTtxr), fj 8e xo 
Otto [xfjc] 8Xr]c xfjc xe[ivouar]c xal xfjc exxoc duoXa^pa- 



and DC plus the (square) on FB is equal to the (sum 
of the squares) on FB and BD. Let the (square) on 
FB have been subtracted from both. Thus, the remain- 
ing (rectangle contained) by AD and DC is equal to the 
(square) on the tangent DB. 

And so let DC A not be through the center of cir- 
cle ABC, and let the center E have been found, and 
let EF have been drawn from E, perpendicular to AC 
[Prop. 1.12]. And let EB, EC, and ED have been joined. 
(Angle) EBD (is) thus a right-angle [Prop. 3.18]. And 
since some straight-line, EF, through the center, cuts 
some (other) straight-line, AC, not through the center, 
at right-angles, it also cuts it in half [Prop. 3.3]. Thus, 
AF is equal to FC. And since the straight-line AC is cut 
in half at point F, let CD have been added to it. Thus, the 
(rectangle contained) by AD and DC plus the (square) 
on FC is equal to the (square) on FD [Prop. 2.6]. Let 
the (square) on FE have been added to both. Thus, the 
(rectangle contained) by AD and DC plus the (sum of 
the squares) on CF and FE is equal to the (sum of the 
squares) on FD and FE. But the (square) on EC is equal 
to the (sum of the squares) on CF and FE. For [angle] 
EFC [is] a right-angle [Prop. 1.47]. And the (square) 
on ED is equal to the (sum of the squares) on DF and 
FE [Prop. 1.47]. Thus, the (rectangle contained) by AD 
and DC plus the (square) on EC is equal to the (square) 
on ED. And EC (is) equal to EB. Thus, the (rectan- 
gle contained) by AD and DC plus the (square) on EB 
is equal to the (square) on ED. And the (sum of the 
squares) on EB and BD is equal to the (square) on ED. 
For EBD (is) a right-angle [Prop. 1.47]. Thus, the (rect- 
angle contained) by AD and DC plus the (square) on 
EB is equal to the (sum of the squares) on EB and BD. 
Let the (square) on EB have been subtracted from both. 
Thus, the remaining (rectangle contained) by AD and 
DC is equal to the (square) on BD. 

Thus, if some point is taken outside a circle, and two 
straight-lines radiate from it towards the circle, and (one) 
of them cuts the circle, and (the other) touches (it), then 
the (rectangle contained) by the whole (straight-line) 
cutting (the circle), and the (part of it) cut off outside 
(the circle), between the point and the convex circumfer- 
ence, will be equal to the square on the tangent (line). 
(Which is) the very thing it was required to show. 

Proposition 37 

If some point is taken outside a circle, and two 
straight-lines radiate from the point towards the circle, 
and one of them cuts the circle, and the (other) meets 
(it), and the (rectangle contained) by the whole (straight- 



106 



ETOIXEIfiN y'. 



ELEMENTS BOOK 3 



\onevr)c, [isxa^u xoO te arpdou xal xfjc xupxfjc; Tiepicpepdac; 
I'aov xfi duo xfjc; upoaumxouarjc;, f\ TtpocnuTtxouaa ecpdcjjexai 
xoO xuxXou. 




KuxXou yap xoO ABr riXfjcp'StL) xi ar^eiov exxo<; xo A, 
xal ooto xoO A 7ip6<; xov ABr xuxXov TtpocnuTixexcoaav 860 
eu-deiai ai ArA, AB, xal rj \iev ATA xe^ivexw xov xuxXov, f] 
8e AB Tipoaumxexw, eaxco 8e xo Otto xCSv AA, Ar laov tu 
dira xfj? AB. Xeyto, oxi r) AB eqxbtxexai xou ABr xuxXou. 

"H)cdw yap xou ABr ecpajixo^ievr) f] AE, xal dAr|cp'dco xo 
xevxpov xoO ABr xuxXou, xal eaxw xo Z, xal £Tie^£U)cdwaav 
ai ZE, ZB, ZA. f) dpa utto ZEA op-dr] eaxiv. xal end f) AE 
scpotTixexai xou ABr xuxXou, xejivei 8s f) ArA, xo dpa Otto 
xfiv A A, Ar I'oov soxl x« aTio xfjc; AE. rjv 8s xal xo Otto 
xfiv A A, Ar taov xG dno xfjc; AB' xo dpa duo xfjc; AE 
I'oov eoxl xG duo xfjc; AB' Tar) dpa f] AE xfj AB. eoxl 8e 
xal f) ZE xfj ZB lor\- 860 8f) ai AE, EZ 860 xau; AB, BZ 
i'oai eiaiv xal pdaig auxaiv xoivf) f) ZA- ywvia dpa f] Otto 
AEZ ywvia xfj utto ABZ saxiv for), op-df] Ss f) Otto AEZ- 
6pi9r) dpa xal f) utio ABZ. xai eoxiv rj ZB sxpaXXo^evr) 
Sidjiexpoc;- f) 8e xfj 8ia[iexpcp xou xuxXou Ttpoc; opiJac; an' 
dxpac; dyo^isvr) scpaTtxexai xou xuxXou- f] AB dpa ecpaTtxexai 
xou ABr xuxXou. ojioimc; 8/) Ssix^iqoexai, xdv xo xsvxpov 
Era xfjc; Ar xuy)(dvr]. 

°Edv dpa xuxXou X/]cpi9fj xi a/j^islov sxxog, arco 8s xou 
ar][ieio\j npbc, xov xuxXov TipooTiiTixwai 860 euiSdai, xal f] 
y.sv auxfiv xspivr) xov xuxXov, f) 8e TipooniTixr), rj 8e xo utio 
oXrjc; xfjc xe[ivouar)<; xal xfjc exxoc; dTioXajipavojisvrjc; ^exac;u 
xou xs af][ieioyj xal xfjc; xupxfjc; Tispicpspdac; Taov xcp aTto 
xfjc; TtpoaTUTtxouarjc;, f] TtpocnuTtxouaa ecpd^sxai xou xuxXou- 
oTiep sSei Sdc;ai. 



line) cutting (the circle), and the (part of it) cut off out- 
side (the circle), between the point and the convex cir- 
cumference, is equal to the (square) on the (straight-line) 
meeting (the circle), then the (straight-line) meeting (the 
circle) will touch the circle. 



D 




For let some point D have been taken outside circle 
ABC, and let two straight-lines, DC A and DB, radiate 
from D towards circle ABC, and let DC A cut the circle, 
and let DB meet (the circle) . And let the (rectangle con- 
tained) by AD and DC be equal to the (square) on DB. 
I say that DB touches circle ABC. 

For let DE have been drawn touching ABC [Prop. 
3.17], and let the center of the circle ABC have been 
found, and let it be (at) F. And let FE, FB, and FD 
have been joined. (Angle) FED is thus a right-angle 
[Prop. 3.18]. And since DE touches circle ABC, and 
DC A cuts (it), the (rectangle contained) by AD and DC 
is thus equal to the (square) on DE [Prop. 3.36]. And the 
(rectangle contained) by AD and DC was also equal to 
the (square) on DB. Thus, the (square) on DE is equal 
to the (square) on DB. Thus, DE (is) equal to DB. And 
FE is also equal to FB. So the two (straight-lines) DE, 
EF are equal to the two (straight-lines) DB, BF (re- 
spectively) . And their base, FD, is common. Thus, angle 
DEF is equal to angle DBF [Prop. 1.8]. And DEF (is) 
a right-angle. Thus, DBF (is) also a right-angle. And 
FB produced is a diameter, And a (straight-line) drawn 
at right-angles to a diameter of a circle, at its extremity, 
touches the circle [Prop. 3.16 corr.]. Thus, DB touches 
circle ABC. Similarly, (the same thing) can be shown, 
even if the center happens to be on AC. 

Thus, if some point is taken outside a circle, and two 
straight-lines radiate from the point towards the circle, 
and one of them cuts the circle, and the (other) meets 
(it), and the (rectangle contained) by the whole (straight- 
line) cutting (the circle), and the (part of it) cut off out- 
side (the circle), between the point and the convex cir- 
cumference, is equal to the (square) on the (straight-line) 
meeting (the circle), then the (straight-line) meeting (the 
circle) will touch the circle. (Which is) the very thing it 



107 



ETOIXEiQN y\ ELEMENTS BOOK 3 

was required to show. 



108 



ELEMENTS BOOK 4 

Construction of Rectilinear Figures In and 

Around Circles 



109 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



"Opoi. 

a'. Sx^K a £ui}uypa|ji[iov etc; a/f^oc eMuypa^ov eyypdcp- 
ea-&ai Xeyexai, oxav exdaxr] x£>v xou eyypacpo^ievou axr^ax- 
oc, yiovifiv exdaxrjc; TiXeupac; tou, etc; o eyypdcpexai, dTtxrjxai. 

P'. S)(fi(jia 8e 6|jioi«c; Ttepl oxrjjia Ttepiypdcpecrdai Xeyexai, 
oxav exdaxr] TiXeupa xoO Tiepiypacpojievou exdaxrjc; ywviac 
tou, tie pi 8 Tiepiypdcpexai, cbixrjxai. 

y'. E/rjiia eMuypajjijiov etc; xuxXov eyypdcpecrdaa Xeyexai, 
oxav exdaxr) ywvia xou eyypacpojievou aTixrjxai xrjc; xou 
xuxXou Tiepicpepeiac;. 

8'. S/'^l jla Se eu-duypa^iov uepl xuxXov Ttepiypdcpe- 
a-Qai Xeyexai, oxav exdaxr) TiXeupa xou Tiepiypacpojievou 
ecpaTix/jxai xrjc; xou xuxXou Ttepicpepeiac;. 

e'. KuxXog 8e etc; a^/j^a 6\±oiu>q eyypdcpeo'dai Xeyexai, 
oxav f) xou xuxXou Ttepicpepeia exdaxrjc; TiXeupac; xou, etc; o 
eyypdcpexai, aTix/jxai. 

<?'. KuxXoc 8e Tiepl axrjjia Tiepiypdcpecdai Xeyexai, oxav 
f) xou xuxXou Tiepicpepeia exdaxrjc; ywviac; xou, uepl o Tie- 
piypdcpexai, aTixrjxai. 

Eu-deTa etc; xuxXov evapjio^ecrdai Xeyexai, oxav xa 
nepaxa auxrjg era xrjc; Tiepicpepeiac; fj xou xuxXou. 



a . 

Etc; xov 8oi3evxa xuxXov xfj Soi5eiarj eu-deia \lt\ [isiZovi 
ouar] xrjc; xou xuxXou Biajiexpou larjv eui&elav evapjioaai. 

A 




"Eaxw 6 Bo-delc; xuxXoc 6 ABr, f) 8e SoTJelaa euiSeTa ^tf) 
(jiei^wv xrjc xou xuxXou 8ia[iexpou f] A. 8el 8f) etc xov ABr 
xuxXov xfj A eu-deia iar]v eu-delav evap^toaai. 

"H)cda) xou ABr xuxXou 8idpiexpo<; f] Br. et uev ouv tar] 
eaxlv f] Br xfj A, yeyovoc; av eir) xo eTiixax^ev evrpuoaxai 



Definitions 

1. A rectilinear figure is said to be inscribed in 
a(nother) rectilinear figure when the respective angles 
of the inscribed figure touch the respective sides of the 
(figure) in which it is inscribed. 

2. And, similarly, a (rectilinear) figure is said to be cir- 
cumscribed about a(nother rectilinear) figure when the 
respective sides of the circumscribed (figure) touch the 
respective angles of the (figure) about which it is circum- 
scribed. 

3. A rectilinear figure is said to be inscribed in a cir- 
cle when each angle of the inscribed (figure) touches the 
circumference of the circle. 

4. And a rectilinear figure is said to be circumscribed 
about a circle when each side of the circumscribed (fig- 
ure) touches the circumference of the circle. 

5. And, similarly, a circle is said to be inscribed in a 
(rectilinear) figure when the circumference of the circle 
touches each side of the (figure) in which it is inscribed. 

6. And a circle is said to be circumscribed about a 
rectilinear (figure) when the circumference of the circle 
touches each angle of the (figure) about which it is cir- 
cumscribed. 

7. A straight-line is said to be inserted into a circle 
when its extemities are on the circumference of the circle. 

Proposition 1 

To insert a straight-line equal to a given straight-line 
into a circle, (the latter straight-line) not being greater 
than the diameter of the circle. 

D 




Let ABC be the given circle, and D the given straight- 
line (which is) not greater than the diameter of the cir- 
cle. So it is required to insert a straight-line, equal to the 
straight-line D, into the circle ABC. 

Let a diameter BC of circle ABC have been drawn.t 



110 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



yap zlc, xov ABr xuxXov xfj A sinJsia Tor) f] BY. zl Bs (isi^cov 
soxlv f) Br xfjc A, xslo'dw xfj A lor] f] TE, xod xsvxpw 
xto T 8iaoxf|[iaxi 8s xc5 TE xuxXoc; ysypdcpiSio 6 EAZ, xod 
£Tie^£U)cdw f) TA. 

'End ouv to T ar^slov xsvxpov soxl xoO EAZ xuxXou, 
far) soxlv f] TA xfj TE. dXXd xfj A f] TE soxiv for) - xal f) A 
apa xfj TA soxiv Tor). 

Etc; apa xov Bo'dsvxa xuxXov xov ABr xfj Bo-dslor] sui5sia 
xfj A tor] Evrpuooxai f] TA- oTisp sSsi Ttoifjoai. 



Therefore, if £?C is equal to D then that (which) was 
prescribed has taken place. For the (straight-line) BC, 
equal to the straight-line D, has been inserted into the 
circle ABC. And if BC is greater than D then let CE be 
made equal to D [Prop. 1.3], and let the circle EAF have 
been drawn with center C and radius CE. And let CA 
have been joined. 

Therefore, since the point C is the center of circle 
EAF, CA is equal to CE. But, CE is equal to D. Thus, 
D is also equal to CA. 

Thus, CA, equal to the given straight-line D, has been 
inserted into the given circle ABC. (Which is) the very 
thing it was required to do. 



Presumably, by finding the center of the circle [Prop. 3.1], and then drawing a line through it. 



P'- 

Etc xov 8oi9svxa xuxXov xw Scdevxi xpiycovcp Eooycoviov 
xptyovov syypd(|>ai. 





'Eoxco 6 So'delc; xuxXoc; 6 ABr, xo 8s 8oif)sv xpiywvov 
xo AEZ- 8sT 5f] zlc, xov ABr xuxXov xw AEZ xpiywvcp 
looyoviov xplyovov eyypd^ai. 

"H)cda) xou ABr xuxXou scpaTixo^iivr] f] H0 xaxd xo A, 
xal ouvsoxdxco npbc, xfj A9 sO'dsia xal xw Ttpoc; auxfj or^slo 
tu A xfj utio AEZ ywvia Tar) f] utio 0Ar, Ttpoc 8s xfj AH 
sui5sia xal xa> Tipoc auxfj or)u.s(cp xfi A xfj utio AZE [ycovia] 
Tor] f] utio HAB, xal etis^sux'&co f) BE 

'End ouv xuxXou xou ABT scpdTtxsxal xic su-dsla f) A0, 
xal duo xrjc xaxd xo A STiacprjc sic; xov xuxXov Birjxxai sO'dsia 
f) Ar, f) apa utio 9Ar Tor) soxl xfj sv x£5 svaXXdi; xou xuxXou 
x^f][iaxi ywvia xfj utio ABr. dXX' rj utio 0Ar xfj utio AEZ 
soxiv lor]' xal r) utio ABr apa ywvia xfj utio AEZ soxiv 
Tor). Bid xd auxa 8f) xal f] utio ArB xfj utio AZE soxiv 
lor]- xal XoiTif] apa f) utio BAr Xomfj xfj utio EAZ soxiv lor) 
[looywviov apa eoxl xo ABr xpiywvov ifi AEZ xpiywvw, 
xal syysypanxai sic tov ABT xuxXov]. 

Etc xov Bo'dsvxa apa xuxXov ifi 8o , f)svxi xpiycovcp 
looyoviov xpiycovov syysypaiixai- OTisp sBsi Tioifjaai. 



Proposition 2 

To inscribe a triangle, equiangular with a given trian- 
gle, in a given circle. 

^\ E 





Let ABC be the given circle, and DEF the given tri- 
angle. So it is required to inscribe a triangle, equiangular 
with triangle DEF, in circle ABC. 

Let GH have been drawn touching circle ABC at A.^ 
And let (angle) HAC, equal to angle DEF, have been 
constructed on the straight-line AH at the point A on it, 
and (angle) GAB, equal to [angle] DFE, on the straight- 
line AG at the point A on it [Prop. 1.23]. And let BC 
have been joined. 

Therefore, since some straight-line AH touches the 
circle ABC, and the straight-line AC has been drawn 
across (the circle) from the point of contact A, (angle) 
HAC is thus equal to the angle ABC in the alternate 
segment of the circle [Prop. 3.32]. But, HAC is equal to 
DEF. Thus, angle ABC is also equal to DEF. So, for the 
same (reasons), ACB is also equal to DFE. Thus, the re- 
maining (angle) BAG is equal to the remaining (angle) 
EDF [Prop. 1.32]. [Thus, triangle ABC is equiangu- 
lar with triangle DEF, and has been inscribed in circle 



111 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



t See the footnote to Prop. 3.34. 

y'- 

Ilepl xov Bo-devxa xuxXov iu Bo'devxi xpiyovo iaoyoviov 
xpiywvov Tiepiypd^ai. 




a r n 

"EaTO 6 Scdslc xuxXo<; 6 ABr, to Se So-fJev xpiycovov 
to AEZ- 8eT 5r| Tiepi xov ABr xuxXov iw AEZ xpiycovcp 
iaoycoviov xpiyovov Tiepiypdtjiai.. 

'ExpEpX^a-dw f] EZ eq>' exdxepa xd [ispf] xaxd xd H, 6 
ar^eia, xal eiXr]q)T&to tou ABr xuxXou xevxpov to K, xod 
Bi^jcda), (be; etu)(£v, euiDeTa f\ KB, xai auvsaxdxo Tipoc; xfj 
KB EU'deia xal xG Tipoc; auxfj ar^dw iu K xfj fjisv utio AEH 
yovia Xar] r) utio BKA, xfj 8e utio AZ9 iar) f] utio BKr, xal 
8id xfiv A, B, T ar^eiwv f))fdwoav scpaTixojievai xou ABr 
xuxXou ai AAM, MBN, NEA 

Kal euel eqxbixovxai xou ABr xuxXou ai AM, MN, NA 
xaxd xa A, B, r ar^eia, duo 8e xou K xsvxpou era xd A, B, 
r ar^eia ETieCsuy^evai elolv ai KA, KB, Kr, bp-Qai apa eialv 
ai Tipoc; xolc; A, B, T ar^eioiz ywviai. xal etiei xou AMBK 
xexpauXeupou ai xeaaapec; ycoviai xexpaaiv op-ddic; law. eiaiv, 
£7i£i8r]Tiep xal tic, Suo xpiywva Siaipdxai xo AMBK, xai siaiv 
opiDal ai utio KAM, KBM ywviai, Xomai apa ai utio AKB, 
AMB 8uaiv opiate; i'aai eiaiv. riai Be xai ai utio AEH, 
AEZ Suaiv opiJau; i'aai- ai apa utio AKB, AMB xdic; utio 
AEH, AEZ Taai eiaiv, £>v f] utio AKB xfj utio AEH eaxiv 
lot]- XoiTif) apa r] utio AMB XoiTifj xfj utio AEZ eaxiv Xar). 
6^ioi«<; 8r) Beix^^asxai, oxi xai f] utio ANB xfj utio AZE 
eaxiv Xar}' xai XoiTif) apa f) utio MAN [Xomfj] xfj utio EAZ 
eaxiv Xar). iaoyoviov apa eaxi xo AMN xpiywvov x£> AEZ 
xpiyovw- xai TiepiyeypaTixai Tiepi xov ABr xuxXov. 

nepl xov Bo'devxa apa xuxXov x£> 8oif)evxi xpiywvw 
iaoyoviov xpiyovov TiepiyeypaTixar oiiep e8ei Tioifjaai. 



ABC]. 

Thus, a triangle, equiangular with the given triangle, 
has been inscribed in the given circle. (Which is) the very 
thing it was required to do. 



Proposition 3 

To circumscribe a triangle, equiangular with a given 
triangle, about a given circle. 

H 

7 D 




L C N 

Let ABC be the given circle, and DEF the given tri- 
angle. So it is required to circumscribe a triangle, equian- 
gular with triangle DEF, about circle ABC. 

Let EF have been produced in each direction to 
points G and H. And let the center K of circle ABC 
have been found [Prop. 3.1]. And let the straight-line 
KB have been drawn, at random, across {ABC). And 
let (angle) BKA, equal to angle DEC, have been con- 
structed on the straight-line KB at the point K on it, 
and (angle) BKC, equal to DFH [Prop. 1.23]. And let 
the (straight-lines) LAM, MBN, and NCL have been 
drawn through the points A, B, and C (respectively), 
touching the circle ABC'J 

And since LM, MN, and NL touch circle ABC at 
points A, B, and C (respectively), and KA, KB, and 
KC are joined from the center K to points A, B, and 
C (respectively), the angles at points A, B, and C are 
thus right-angles [Prop. 3.18]. And since the (sum of the) 
four angles of quadrilateral AMBK is equal to four right- 
angles, inasmuch as AMBK (can) also (be) divided into 
two triangles [Prop. 1.32], and angles KAM and KBM 
are (both) right-angles, the (sum of the) remaining (an- 
gles), AKB and AMB, is thus equal to two right-angles. 
And DEC and DEF is also equal to two right-angles 
[Prop. 1.13]. Thus, AKB and AMB is equal to DEC 
and DEF, of which AKB is equal to DEC. Thus, the re- 
mainder AMB is equal to the remainder DEF. So, sim- 
ilarly, it can be shown that LNB is also equal to DFE. 
Thus, the remaining (angle) MLN is also equal to the 



112 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



t See the footnote to Prop. 3.34. 

5'. 

Etc to 6otl>ev Tplycovov xuxXov eryypd^ai. 

A 




T5gto> to SoiSev Tpiywvov to ABE 8a 5r] el? to ABr 
Tpiywvov xuxXov eyypdijiai. 

TeT^nqcrdMaav od Otto ABr, ArB ycoviai 8i)(a Tdig BA, 
TA su-vMaig, xod aujipaXXsTwaav dXXr|Xaig xaTa to A 
o/jjisTov, xal fj/iJioaav duo tou A km Tag AB, Br, EA 
sMelag xdiJeToi al AE, AZ, AH. 

Kod inz\ lot) eotIv f\ uno ABA ytovla T/j utto TBA, 
eaTi 8e xai op'dr) i\ uno BEA op^fj t/j utto BZA lot], 86o 
8r] Tpiycovd eoTi Td EBA, ZBA Tag 80o ycovlag Talc; Sual 
ycoviaig laag e)( 0VTa xal ( i ' av ^Xeupdv [iia TiXsupa Taiqv tt]v 
OnoTEivouaav Otto [ilav twv lacov ywviaiv xoivfjv auTfiv tt]v 
BA- xal Tag XoiTidg apa TcXsupag Tdig XoiTtaTg TcXsupdig laag 
sg'ouaiv Tor) apa f] AE Tfj AZ. Bid Ta ai)Ta 8f) xal f) AH 
tt) AZ eoTiv Tor), al Tpelg apa euifeTai al AE, AZ, AH 
I'oai dXXr]Xaig eloiv 6 apa xsvTpG tG A xal 8iaoTr]jj.aTi evl 
tGv E, Z, H xuxXog ypacpojisvog r^gei xal 8id tcov XoitcSSv 
a/][id«v xal scpdcJjSTai t«v AB, Br, TA euifteiwv 8id to 
op-ddg dvai Tag Tcpog Tolg E, Z, H ar^eion; yioviag. el yap 
T£[iei auTag, scxuai f\ Tfj Bia^STpcp tou xuxXou Ttpog opiDdg 
an' axpag dyojievr) evTog TUTCTOuaa tou xuxXou- onep aTO- 
tiov £8el)cdr)- oux apa 6 xevTpip t£5 A SiarjTrj^aTi Se evl tGv 
E, Z, H ypacpojisvog xuxXog TE^iel Tag AB, Br, TA eu^dag' 
£(pd(|j£Tai apa auTfiv, xal scnai 6 xuxXog eyyeypajijisvog rig 
to ABr Tplytovov. syyeypdcp-dw tbg 6 ZHE. 

Elg apa to 8o$ev Tpiywvov to ABr xuxXog eyyeypaKTai 
6 EZH- oTiep eBsi noifjaaL. 



[remaining] (angle) EDF [Prop. 1.32]. Thus, triangle 
LMN is equiangular with triangle DEF. And it has been 
drawn around circle ABC. 

Thus, a triangle, equiangular with the given triangle, 
has been circumscribed about the given circle. (Which is) 
the very thing it was required to do. 



Proposition 4 

To inscribe a circle in a given triangle. 



A 




Let ABC be the given triangle. So it is required to 
inscribe a circle in triangle ABC. 

Let the angles ABC and ACB have been cut in half by 
the straight-lines BD and CD (respectively) [Prop. 1.9], 
and let them meet one another at point D, and let DE, 
DF, and DC have been drawn from point D, perpendic- 
ular to the straight-lines AB, BC, and CA (respectively) 
[Prop. 1.12]. 

And since angle ABD is equal to CBD, and the right- 
angle BED is also equal to the right-angle BFD, EBD 
and FBD are thus two triangles having two angles equal 
to two angles, and one side equal to one side — the (one) 
subtending one of the equal angles (which is) common to 
the (triangles) — (namely), BD. Thus, they will also have 
the remaining sides equal to the (corresponding) remain- 
ing sides [Prop. 1.26]. Thus, DE (is) equal to DF. So, 
for the same (reasons), DC is also equal to DF. Thus, 
the three straight-lines DE, DF, and DC are equal to 
one another. Thus, the circle drawn with center D, and 
radius one of E, F, or Gj will also go through the re- 
maining points, and will touch the straight-lines AB, BC, 
and CA, on account of the angles at E, F, and G being 
right-angles. For if it cuts (one of) them then it will be 
a (straight-line) drawn at right-angles to a diameter of 
the circle, from its extremity, falling inside the circle. The 
very thing was shown (to be) absurd [Prop. 3.16]. Thus, 
the circle drawn with center D, and radius one of E, F, 



113 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



or G, does not cut the straight-lines AB, BC, and CA. 
Thus, it will touch them and will be the circle inscribed 
in triangle ABC. Let it have been (so) inscribed, like 
FGE (in the figure). 

Thus, the circle EFG has been inscribed in the given 
triangle ABC. (Which is) the very thing it was required 
to do. 



t Here, and in the following propositions, it is understood that the radius is actually one of DE, DF, or DG. 



e . 

Ilepi to Bo'dev Tpiycovov xuxXov 7iepiypd<j>ai. 




"Eaxw to SoTJev xpiywvov to ABE 8a Se rcspl to Bo'dev 
Tpiycovov to ABT xuxXov 7tepiypd(];ai. 

TeT^iiqo'Owoav ai AB, Ar euifteTai Si)(a xaTa Ta A, E 
ar^ela, xai arco tcov A, E arjueitov xau; AB, AT Tipoc. 6pi9dc; 
f])Cdwoav ai AZ, EZ- aupmeaouvTai 5r) f]Toi evTo<; toO ABr 
Tpiycovou f] em Trj<; Br eO'deiac. fj cxtoc. Trjc. Br. 

Sujj.7ii7iT£TCdaav upoTEpov Ivtoc. xaTa to Z, xai £Tie^£U)cd- 
coaav ai ZB, ZT, ZA. xai end Tar) eaTiv f) AA Tfj AB, xoivr) 
8e xai Ttpoc. opiSdc. f] AZ, pdai<; dpa f] AZ pdaei Tfj ZB eaTiv 
lor\. b[io'ux>z 8t) Bei^o^iev, oti xai f] TZ T/j AZ ecrciv tar]- 
6aie xai f] ZB Tfj ZT eaTiv Tar]- ai TpeX? dpa ai ZA, ZB, ZT 
laai dXXf|Xaic. eiaiv. 6 apa xevTpco tw Z 8iaaTf|[iaTi 8e evi 
twv A, B, r xvxkoc, ypacpojievoc; fj^ei xai 8ia tcov XoitcSSv 
ar^eicov, xai eaTai 7iepiyeypa[i(jievoc. 6 xuxXoc; Tiepi to ABr 
Tpiywvov. Tiepiyeypdcp-dco cbc. 6 ABE 

AXXa 8r) ai AZ, EZ aupmmTSTwaav iiu Tfjc. BT eu-Mac. 
xaTa to Z, cbc; ex £l ^ BeuTepac. xaTaypacpfjc,, xai 
CTieCeux'dM ^ AZ. 6[ioicoc 8f] 8eic;o[iev, oti to Z aruielov 
xevTpov eaTi tou Ttepi to ABT Tpiycovov Ttepiypacpo^ievou 
xuxXou. 

AXXa 8r] ai AZ, EZ aujjiTimTSTwaav cxtoc. toO ABT 
Tpiycivou xaTa to Z TtdXiv, cbc. l/ei em Trjc. TpiTr]C xaTa- 
ypacprjc,, xai eTte£eu)cd«aav ai AZ, BZ, TZ. xai eizsl TtdXiv 
lot] eaTiv f] AA Tfj AB, xoivf) Be xai Tipoc, op'dac, f] AZ, p&au; 
dpa f] AZ pdaei xfj BZ eaTiv i'ar). ojioimc, 8f] Sei^o^iev, oti 
xai f] TZ Tfj AZ eaTiv i'ar)- waTe xai f] BZ Tfj ZT eaTiv for) - 
6 dpa [rcdXiv] xevTpw t£> Z SiaaT^aTi 8e evi itov ZA, ZB, 
ZT xuxXoc ypacpo^ievoc ffesi xai 8id t£Sv Xoitiwv ar^eiwv, 
xai eaTai Ttepiyeypa[i^evoc Ttepi to ABT Tpiycovov. 

Ilepl to Bo-dev apa Tpiywvov xuxXo^ uepiyeypaKTar 
OTiep e8ei Koifjaai. 



Proposition 5 



To circumscribe a circle about a given triangle. 




Let ABC be the given triangle. So it is required to 
circumscribe a circle about the given triangle ABC. 

Let the straight-lines AB and AC have been cut in 
half at points D and E (respectively) [Prop. 1.10]. And 
let DF and EF have been drawn from points D and E, 
at right-angles to AB and AC (respectively) [Prop. 1.11]. 
So (DF and EF) will surely either meet inside triangle 
ABC, on the straight-line BC, or beyond BC. 

Let them, first of all, meet inside (triangle ABC) at 
(point) F, and let FB, FC, and FA have been joined. 
And since AD is equal to DB, and DF is common and 
at right-angles, the base AF is thus equal to the base FB 
[Prop. 1.4]. So, similarly, we can show that CF is also 
equal to AF. So that FB is also equal to FC. Thus, the 
three (straight-lines) FA, FB, and FC are equal to one 
another. Thus, the circle drawn with center F, and radius 
one of A, B, or C, will also go through the remaining 
points. And the circle will have been circumscribed about 
triangle ABC. Let it have been (so) circumscribed, like 
ABC (in the first diagram from the left) . 

And so, let DF and EF meet on the straight-line 
BC at (point) F, like in the second diagram (from the 
left). And let AF have been joined. So, similarly, we can 
show that point F is the center of the circle circumscribed 
about triangle ABC. 

And so, let DF and EF meet outside triangle ABC, 
again at (point) F, like in the third diagram (from the 
left). And let AF, BF, and CF have been joined. And, 
again, since AD is equal to DB, and DF is common and 
at right-angles, the base AF is thus equal to the base BF 
[Prop. 1.4]. So, similarly, we can show that CF is also 
equal to AF. So that BF is also equal to FC. Thus, 



114 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



[again] the circle drawn with center F, and radius one 
of FA, FB, and FC, will also go through the remaining 
points. And it will have been circumscribed about trian- 
gle ABC. 

Thus, a circle has been circumscribed about the given 
triangle. (Which is) the very thing it was required to do. 



Proposition 6 



Etc; xov Boi9svxa xuxXov xexpdywvov eyypdijiai. 

A 



To inscribe a square in a given circle. 

A 




"Eaxw f] Bo'vteu; xuxXoc; 6 ABrA- Bel 8r) zlc. xov ABrA 
xuxXov xsxpdyiovov eyypdtjiai. 

"H)(Tf)cL)acxv xou ABrA xuxXou Buo Bidjiexpoi npbc, opiSdc; 
dXXfjXaic; ad Ar, BA, xod eiteCeux'dwaav al AB, Br, TA, 
AA. 

Kal em\lar\ eaxlv f] BE xfj EA- xevxpov yap to E- xoivrj 
Be xal Tipoc; 6pi9dc; f] EA, f3dai<; dpa f) AB pdoei xrj AA Tar) 
eaxiv. Sid xd auxd 5r] xal exaxepa x£5v Br, TA exaxepa 
xfiv AB, AA for) eoxiv iooTiXeupov dpa eoxl xo ABrA 
xexpaTiXsrupov. Xsy« 8r], oxi xal 6p$oywviov. inel yap f] 
BA su-dela Sidjisxpoc; eoxi xou ABrA xuxXou, /jjiixuxXiov 
dpa eaxl xo BAA- 6pi3r) dpa #j utto BAA ycovia. Sid xa 
auxd Br] xal exdaxr) xGv utio ABr, BrA, TAA op-Qr] soxiv 
6p$oycoviov dpa eaxl xo ABTA xexpdnXeupov. eBsix'dif) Be 
xal EaonXsupov xsxpdytovov dpa saxiv. xal syysypanxai eic, 
xov ABrA xuxXov. 

Eic; dpa xov 8oi5evxa xuxXov xsxpdywvov syyeyparcxai 
xo ABrA- ojiep eBsi noifjaai. 




Let ABCD be the given circle. So it is required to 
inscribe a square in circle ABCD. 

Let two diameters of circle ABCD, AC and BD, have 
been drawn at right-angles to one another, t And let AB, 
BC, CD, and DA have been joined. 

And since BE is equal to ED, for E (is) the center 
(of the circle), and EA is common and at right-angles, 
the base AB is thus equal to the base AD [Prop. 1.4]. 
So, for the same (reasons), each of BC and CD is equal 
to each of AB and AD. Thus, the quadrilateral ABCD 
is equilateral. So I say that (it is) also right-angled. For 
since the straight-line BD is a diameter of circle ABCD, 
BAD is thus a semi-circle. Thus, angle BAD (is) a right- 
angle [Prop. 3.31]. So, for the same (reasons), (angles) 
ABC, BCD, and CD A are also each right-angles. Thus, 
the quadrilateral ABCD is right-angled. And it was also 
shown (to be) equilateral. Thus, it is a square [Def. 1.22]. 
And it has been inscribed in circle ABCD. 

Thus, the square ABCD has been inscribed in the 
given circle. (Which is) the very thing it was required 
to do. 



t Presumably, by finding the center of the circle [Prop. 3.1], drawing a line through it, and then drawing a second line through it, at right-angles 
to the first [Prop. 1.11]. 



Ilepl xov So-devxa xuxXov xexpdywvov uepiypatjiai. 



Proposition 7 

To circumscribe a square about a given circle. 



115 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



'Eaxw 6 So-del; xuxXo<; 6 ABrA- Bel 8rj Tiepi xov ABrA 
xuxXov xexpdycovov Ttepiypdijiai. 

H A Z 



B 



@ r k 

"HyT&coaav xou ABrA xuxXou 8uo Bid^expoi Tipoc; 6pi9dc; 
dXXf|Xai<; ai Ar, BA, xai Bid iSv A, B, T, A arjjjidcov rj/iiko- 
aav ecpairxo^ievai xoO ABrA xuxXou ai ZH, H6, 6K, KZ. 

'Etc! ouv ecpdrcxexai rj ZH xou ABrA xuxXou, duo 8s 
xou E xevxpou era. xr]v xaxd xo A S7ioccpf)v eiteCeuxxai f] 
EA, ai apa npoc, xw A ywviai op-dai eiaiv. Bid xa auxa 
8rj xai ai irpoi; xoT? B, T, A ar^eiou; ytoviai opiJai eiaiv. 
xai £7te! op-Qi] eaxiv f] (mo AEB ycovia, eaxi Be op'dr] xal rj 
utio EBH, TiapdXX/jXoi; apa eaxiv f] H9 xfj Ar. 8id xd auxa 
8f) xai r) Ar xfj ZK eaxi 7tapdXXr]Xo<;. uoxe xai f) H9 xfj 
ZK eaxi TtapdXXr)Xo<;. ojioiwc; Bf] Bei^o^iev, oxi xal exaxepa 
xwv HZ, 0K xfj BEA eaxi TtapdXXr)Xo<;. 7tapaXXr]X6ypa[ipc 
apa eaxi xa HK, Hr, AK, ZB, BK- i'ar) apa eaxiv f) [lev 
HZ xfj 0K, f) Be H9 xrj ZK. xai ercel iarj eaxiv f) Ar xfj 
BA, dXXa xai f) jiev Ar exaxepa xGv H6, ZK, r] Be BA 
exaxepa xfiv HZ, 9K eaxiv for] [xai exaxepa apa xwv H0, 
ZK exaxepa x£>v HZ, 0K eaxiv for]], iaouXeupov apa eaxi xo 
ZH0K xexpduXeupov. Xeyco Br), oxi xai op-doywviov. enel 
yap TtapaXXr)X6ypa^6v eaxi xo HBEA, xai eaxiv op'df] f) 
utio AEB, op-df) apa xal f] utio AHB. 6[io[w<; Br] Bei^oiJiev, 
oxi xai ai npbz toiz O, K, Z ywviai op'dai eiaiv. 6p , doy«viov 
apa eaxi xo ZH6K. eBeix^r] Be xai iaoTiXeupov xexpdyovov 
apa eaxiv. xal TtepiyeypaTtxai uepl xov ABrA xuxXov. 

nepl xov So-devxa apa xuxXov xexpdyovov TtepiyeypaTtxai- 
OTiep e8ei Ttoirjaai. 



Let ABCD be the given circle. So it is required to 
circumscribe a square about circle ABCD. 

G A 



B 



D 



H C K 

Let two diameters of circle ABCD, AC and BD, have 
been drawn at right-angles to one another. ^ And let FG, 
GH, HK, and KF have been drawn through points A, 
B, C, and D (respectively), touching circle ABCD} 

Therefore, since FG touches circle ABCD, and EA 
has been joined from the center E to the point of contact 
A, the angles at A are thus right-angles [Prop. 3.18]. So, 
for the same (reasons), the angles at points B, C, and 
D are also right-angles. And since angle AEB is a right- 
angle, and EBG is also a right-angle, GH is thus parallel 
to AC [Prop. 1.29]. So, for the same (reasons), AC is 
also parallel to FK. So that GH is also parallel to FK 
[Prop. 1.30]. So, similarly, we can show that GF and 
HK are each parallel to BED. Thus, GK, GC, AK, FB, 
and BK are (all) parallelograms. Thus, GF is equal to 
HK, and GH to FK [Prop. 1.34]. And since AC is equal 
to BD, but AC (is) also (equal) to each of GH and FK, 
and BD is equal to each of GF and HK [Prop. 1.34] 
[and each of GH and FK is thus equal to each of GF 
and HK], the quadrilateral FGHK is thus equilateral. 
So I say that (it is) also right-angled. For since GBEA 
is a parallelogram, and AEB is a right-angle, AGB is 
thus also a right-angle [Prop. 1.34]. So, similarly, we can 
show that the angles at H, K, and F are also right-angles. 
Thus, FGHK is right-angled. And it was also shown (to 
be) equilateral. Thus, it is a square [Def. 1.22]. And it 
has been circumscribed about circle ABCD. 

Thus, a square has been circumscribed about the 
given circle. (Which is) the very thing it was required 
to do. 



t See the footnote to the previous proposition. 
* See the footnote to Prop. 3.34. 



116 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



Etc to 6oi9£v TeTpdywvov xuxXov syYpd^ai. 
"Eaxco to BotJev xsxpdycovov to ABrA. 8a 5r| sic; to 
ABrA TETpdywvov xuxXov eyYpd^ai. 

A E A 



K 



B 

TeT[if]o , d« exaTepa tcov AA, AB 8i)(a xaTa Ta E, Z 
arista, xal Sid (lev tou E ouoTspa tcov AB, TA TiapdXXrjXoc; 
r^/iSio 6 E9, 8id 8s tou Z orcoTspa xwv AA, Br TtapdXXrjXoc; 
fj/iSw f) ZK' uapaXXr)X6Ypa|ji[j.ov dpa eaTiv exacrcov xSv 
AK, KB, A9, 6A, AH, Hr, BH, HA, xod ai aTtervavTiov 
ai)TG)v TiXeupal StjXovoti I'aai [eiaiv]. xal snel lar\ sgtIv f) 
AA Tfj AB, xa[ ecrci xfj<; [lev AA r^iiaeia f] AE, Tfjc; 6e AB 
f\[iiaeiaL f\ AZ, lar\ dpa xal f\ AE Tfj AZ' wots xal ai am- 
vavTiov i'ar) dpa xal f\ ZH Tfj HE. 6^io[m<; 8r] Bd^o^ierv, oti 
xal exaTepa t£>v H0, HK exaTepa twv ZH, HE eaTiv Xary 
ai Teaaapsc dpa ai HE, HZ, H0, HK laai dXXiqXaii; [eiaiv]. 
6 dpa xevTpo \ism tw H SiaaT^aTi 8e evl xuv E, Z, 0, K 
xuxXoc; YP°tcp6^ievo<; fj^ei xal 8id tGv Xomwv ar^eiwv xal 
ecpdijiSTai twv AB, Br, TA, A A euiDeiwv 8id to opiDdt; eivai 
Tag npbc, Toiz E, Z, <d, K Y^viac;' ei yap tejisI 6 xuxXoc; Tag 
AB, Br, TA, AA, f] Tfj 8ia|jieTpcp tou xuxXou rcpoc; opi&dc; 
caz dxpac; dyoiievr) evto<; TieaELTai tou xuxXou' onep aTOTtov 
sSei/i}/). oux dpa 6 xevTptp tG H BiaaTr^aTi 8s fevi iwv E, 
Z, 0, K xuxXog ypacpojisvog tsjiei Tag AB, Br, TA, AA 
su-deiag. ecpd(|ieTai dpa auT&v xal eaTai eYYSYpa\i\ievoci si<z 
to ABrA TETpaYWvov. 

Ei<; dpa to Bo-dev TeTpdyMvov xuxXog EYY£YP a7txal ' 
onep s8ei Ttoirjaai. 



ft'. 

nspi to Bo'dev T£TpdY«vov xuxXov nepiYpd^ai. 
TiaTCO to Boftsv TCTpaYwvov to ABrA- BsT Br) xcepl to 
ABrA TSTpdYWvov xuxXov nxpiYpd^ai. 



Proposition 8 

To inscribe a circle in a given square. 
Let the given square be ABCD. So it is required to 
inscribe a circle in square ABCD. 

A E D 



K 



B H C 

Let AD and AB each have been cut in half at points E 
and F (respectively) [Prop. 1.10]. And let EH have been 
drawn through E, parallel to either of AB or CD, and let 
FK have been drawn through F, parallel to either of AD 
or BC [Prop. 1.31]. Thus, AK, KB, AH, HD, AG, GC, 
BG, and GD are each parallelograms, and their opposite 
sides [are] manifestly equal [Prop. 1.34]. And since AD 
is equal to AB, and AE is half of AD, and AF half of 
AB, AE (is) thus also equal to AF. So that the opposite 
(sides are) also (equal). Thus, FG (is) also equal to GE. 
So, similarly, we can also show that each of GH and GK 
is equal to each of FG and GE. Thus, the four (straight- 
lines) GE, GF, GH, and GK [are] equal to one another. 
Thus, the circle drawn with center G, and radius one of 
E, F, H, or K, will also go through the remaining points. 
And it will touch the straight-lines AB, BC, CD, and 
DA, on account of the angles at E, F, H, and K being 
right-angles. For if the circle cuts AB, BC, CD, or DA, 
then a (straight-line) drawn at right-angles to a diameter 
of the circle, from its extremity, will fall inside the circle. 
The very thing was shown (to be) absurd [Prop. 3.16]. 
Thus, the circle drawn with center G, and radius one of 
E, F, H, or K, does not cut the straight-lines AB, BC, 
CD, or DA. Thus, it will touch them, and will have been 
inscribed in the square ABCD. 

Thus, a circle has been inscribed in the given square. 
(Which is) the very thing it was required to do. 

Proposition 9 

To circumscribe a circle about a given square. 
Let ABCD be the given square. So it is required to 
circumscribe a circle about square ABCD. 



117 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



'ETU^eux'detaai yap ai Ar, BA xe^ivexcoaav dXXr|Xa<; 
xaxa to E. 



A 




Kai end Tar) eaxlv f| AA xfj AB, xoivf] Se f) Ar, 860 
Sf] ai AA, Ar Sua! xdi<; BA, Ar I'aai eiaiv xal pdau; f) 
Ar pdaei xfj Br Xor\- ywvia dpa f] uko AAr yovia xfj utio 
BAr Tar] laxiv f] dpa 0tc6 AAB ytovia Bi^a xex^trjxai utio 
xfjc; Ar. 6[io[«<; 8f) 8eie;ouev, 6x1 xal exdaxr] xc5v Gtco ABr, 
BrA, TAA 5ixa xex^rjxai utio xfiv Ar, AB eMeiCSv. xal 
CTtel iar] eaxlv f) utio AAB ywvia xfj utio ABr, xa( eaxi 
xfj? uev utio AAB f)uiaeia f) utio EAB, xfjc; Se 0tc6 ABr 
f]\j.iaeia r] utio EBA, xal f) Gtco EAB dpa xfj utio EBA eaxiv 
ten)' £>axe xal TiXeupa f) EA xfj EB eaxiv lay). b[io'ux>c, Sf] 
8eic;o[iev, 6x1 xal exaxepa xwv EA, EB [euT!)eiGv] exaxepa 
x£Sv Er, EA lay] eaxiv. ai xeaaapec; dpa ai EA, EB, Er, 
EA Taai dXXrjXaic; eiaiv. 6 dpa xevxpw ifi E xal Siaaxrj^taxi 
evl iwv A, B, r, A xuxXoc; ypacpojievoc; fjc;ei xal Sid xwv 
Xoitiwv arj^ieiwv xal eaxai Tiepiyeypa[i[ievoc; Ttepl xo ABrA 
xexpdywvov. Tiepiyeypdcp'dw «<; 6 ABrA. 

nepl xo 8oi9ev dpa xexpdywvov xuxXoc TiepiyeypaTixar 
OTtep e8ei Tioirjaai. 



l . 

'IaoaxeXec; xpiywvov auaxf|aaaiL>ai e^ov exaxepav xwv 
Tcpog xfj pdaei ywvifiv 8iTiXaa[ova xfjc; Xomfj^ . 

'Exxeia'dw xu; euiJeTa f) AB, xal xex[if]adw xaxa xo 
r a/][ieIov, waxe xo utio xwv AB, Br Tiepiexouevov 
opiDoywviov laov eTvai x£> arco xfjc; TA xexpaycovw' xal 
xevxpo xc5 A xal 8iaaxf|[iaxi xw AB xuxXoc; yeypdcpif)w 
6 BAE, xal evr)p[i6adw eic; xov BAE xuxXov xfj Ar eMeia 
[ly) [xeiCovi ouarj xfjc; xou BAE xuxXou 8ia[iexpou lot] eu-dela 
f] BA- xal CTieCeux'dwaav ai AA, Ar, xal Tiepiyeypdcp'dw 
Ttepl xo ArA xpiywvov xuxXoc; 6 ArA. 



AC and BD being joined, let them cut one another at 

E. 



A 





E 









c 



And since DA is equal to AB, and AC (is) common, 
the two (straight-lines) DA, AC are thus equal to the two 
(straight-lines) BA, AC. And the base DC (is) equal to 
the base BC. Thus, angle DAC is equal to angle BAC 
[Prop. 1.8]. Thus, the angle DAB has been cut in half 
by AC. So, similarly, we can show that ABC, BCD, and 
CD A have each been cut in half by the straight-lines AC 
and DB. And since angle DAB is equal to ABC, and 
EAB is half of DAB, and EBA half of ABC, EAB is 
thus also equal to EBA. So that side EA is also equal 
to EB [Prop. 1.6]. So, similarly, we can show that each 
of the [straight-lines] EA and EB are also equal to each 
of EC and ED. Thus, the four (straight-lines) EA, EB, 
EC, and ED are equal to one another. Thus, the circle 
drawn with center E, and radius one of A, B, C, or D, 
will also go through the remaining points, and will have 
been circumscribed about the square ABCD. Let it have 
been (so) circumscribed, like ABCD (in the figure). 

Thus, a circle has been circumscribed about the given 
square. (Which is) the very thing it was required to do. 

Proposition 10 

To construct an isosceles triangle having each of the 
angles at the base double the remaining (angle) . 

Let some straight-line AB be taken, and let it have 
been cut at point C so that the rectangle contained by 
AB and BC is equal to the square on CA [Prop. 2.11]. 
And let the circle BDE have been drawn with center A, 
and radius AB. And let the straight-line BD, equal to 
the straight-line AC, being not greater than the diame- 
ter of circle BDE, have been inserted into circle BDE 
[Prop. 4.1]. And let AD and DC have been joined. And 
let the circle ACD have been circumscribed about trian- 
gle ACD [Prop. 4.5]. 



118 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 




Kal etceI to utco x65v AB, Br Taov eaxi x£> duo xfjc 
AT, iar] 8s f] Ar xrj BA, to dpa utio xuv AB, Br taov 
eaxl xc5 dn:6 xfj<; BA. xod ercel xuxXou xou ArA ei'Xr]Tcxa[ xi 
ar^ieTov exxoc; xo B, xal dmo xou B Ttp6<; xov ArA xuxXov 
TcpoaTccTcxtoxaai 8uo eu-delai ai BA, BA, xal f] [lev auxcov 
xe^vei, f) Be: TtpoaTimxei, xa[ eaxi xo utco xwv AB, Br Taov 
xw (xko xfjc BA, f] BA apa ecpdacxexai xou ArA xuxXou. CTcei 
ouv ecp&Tcxexai [iev f] BA, duo 8e xrjc xaxd xo A enacprjg 
8irjxxai f) Ar, f] apa Otto BAr ycovid for) eoxl xrj ev xfi 
evaXXdc; xou xuxXou xpir|[iaxi yiovia xrj utio AAr. eTiei ouv 
Iar] eaxiv f) utio BAr xrj utco AAr, xoivr] Tcpoaxeia'fko rj 
Otto TAA- 6Xr) apa f) utio BAA Tar] eaxl 8ual xau; utio TAA, 
AAr. dXXa xalg Otio TAA, AAr lot] eaxiv f] exxo<; f) utio 
BTA- xal f] Otto BAA apa iar) eaxl xrj utio BrA. dXXa rj 
Otto BAA xrj Otto TBA eaxiv lor), end xal TiXeupa rj AA 
xrj AB eaxiv 'iar}- uaxs xal f] Otio ABA xrj Otio BrA eaxiv 
far), ai xpdc apa ai Otio BAA, ABA, BrA laai dXXr|Xai<; 
eiaiv. xal excel i'ar) saxlv f) utio ABr yovia xrj utio BrA, 
Xar] eaxl xal TtXeupd f] BA TiXeupa xrj Ar. dXXa r) BA xrj 
TA Oiroxeixai iar]- xal f] TA apa xrj TA eaxiv Xaf]- waxe xal 
yovia r] Otto TAA yovia xrj utio AAr eaxiv iar]' ai apa 
utco TAA, AAr xfj<; utio AAr eiai 8iTtXaaiou<;. Xar] Se f] 
utco BrA xau; utco TAA, AAr- xal f] utio BrA apa xfj<; utio 
TAA eaxi BiTcXfj. iar) 8e r] utio BrA exaxepa xwv utio BAA, 
ABA- xal exaxepa apa xwv Otio BAA, ABA xfjc utco AAB 
eaxi BiTcXfj. 

'IaoaxeXet; apa xpiywvov auveaxaxai xo ABA e/ov 
exaxepav xwv Tcpoc; xrj AB pdaei ywvifiv BiTcXaaiova xfj<; 
Xomrjc OTcep e8ei Tioirjaai. 



ia'. 

EE? xov 8o$evxa xuxXov Tievxdywvov iaoTcXeupov xe xal 




And since the (rectangle contained) by AB and BC 
is equal to the (square) on AC, and AC (is) equal to 
BD, the (rectangle contained) by AB and BC is thus 
equal to the (square) on BD. And since some point B 
has been taken outside of circle ACD, and two straight- 
lines BA and BD have radiated from B towards the cir- 
cle ACD, and (one) of them cuts (the circle), and (the 
other) meets (the circle), and the (rectangle contained) 
by AB and BC is equal to the (square) on BD, BD thus 
touches circle ACD [Prop. 3.37]. Therefore, since BD 
touches (the circle), and DC has been drawn across (the 
circle) from the point of contact D, the angle BDC is 
thus equal to the angle DAC in the alternate segment of 
the circle [Prop. 3.32]. Therefore, since BDC is equal 
to DAC, let CD A have been added to both. Thus, the 
whole of BDA is equal to the two (angles) CD A and 
DAC. But, the external (angle) BCD is equal to CD A 
and DAC [Prop. 1.32]. Thus, BDA is also equal to 
BCD. But, BDA is equal to CBD, since the side AD is 
also equal to AB [Prop. 1.5]. So that DBA is also equal 
to BCD. Thus, the three (angles) BDA, DBA, and BCD 
are equal to one another. And since angle DBC is equal 
to BCD, side BD is also equal to side DC [Prop. 1.6]. 
But, BD was assumed (to be) equal to CA. Thus, CA 
is also equal to CD. So that angle CD A is also equal to 
angle DAC [Prop. 1.5]. Thus, CD A and DAC is double 
DAC. But BCD (is) equal to CD A and DAC. Thus, 
BCD is also double CAD. And BCD (is) equal to to 
each of BDA and DBA. Thus, BDA and DBA are each 
double DAB. 

Thus, the isosceles triangle ABD has been con- 
structed having each of the angles at the base BD double 
the remaining (angle). (Which is) the very thing it was 
required to do. 

Proposition 11 
To inscribe an equilateral and equiangular pentagon 



119 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



iaoycoviov eyypdtjiai. 




"Eaxw 6 Bo'dek; xuxXoc; 6 ABrAE- 8eT 8r) ei<; xov 
ABrAE xuxXov Tievxdycovov iaoTtXeupov xe xal iaoycjviov 
eyypdtjiai.. 

'Exxdcrdw xpiytovov iaoaxeXec to ZH0 8inXaaiova 
e)(ov exaxepav xGv Tipoc; xou; H, ywviaiv xrjc Tipog xcp Z, 
xai eyyeypdcpiEko ei<; xov ABrAE xuxXov xfi ZH6 xpiywvw 
iaoyciiviov xpiycovov to AEA, &axe xfj [lev ixpog xto Z yojvia 
Tarjv eTvai xrjv utio TAA, exaxepav 6e xc5v Tipoc; xolc; H, 
i'or)v exaxepa x«v utio ArA, TAA- xal exaxepa dpa xfiv 
utco ArA, TAA xfjc utco TAA eaxi SiTiXrj. xexjnqcdt) 8r| 
exaxepa xfiv utio ArA, TAA Stya Otco exaxepac; x«v TE, 
AB eMeioSv, xal eiceCeux^woav ai AB, Br, AE, EA. 

'Eiiei ouv exaxepa xcov utio ArA, TAA ycoviwv 6l- 
TcXaaicov eaxl xfjc; uico TAA, xal xexjrrjjievai eial 8i)(a utio 
xfiv TE, AB euiSeicov, ai itevxe dpa ycoviai ai utio AAr, 
ArE, EEA, TAB, BAA I'oai dXXf)Xai<; eiaiv. ai 8e laai 
ywviai era latov nepicpepeuov pepiqxaaLv ai nevxe dpa nz- 
picpepeiai ai AB, Br, EA, AE, EA laai dXXrjXaic; eiaiv. utio 
8e xdc laac; Tcepicpepeiac; i'oai euifteTai UTcoxeivouaiv ai itevxe 
dpa euOelai ai AB, Br, TA, AE, EA laai dXXf|Xaic; eiaiv 
iaoTtXeupov dpa eoxi xo ABrAE Tievxdycovov. Xeyto 8iq, 
oxi xal iaoytoviov. enel yap f) AB Tcepicpepeia xfj AE tcs- 
pi9epeia eaxlv Tar), xoivr) Tcpoaxeiai^co f) BrA- oXr) dpa f] 
ABrA nepicpepia 6Xr) xfj EArB nepicpepeia eaxlv i'or). xal 
PePrjxev era ^ev xfjg ABrA nepicpepeiac ycovia f] utco AEA, 
era Se xfjc EArB Tcepicpepeiac; ywvia f] utio BAE- xal f) utio 
BAE dpa yiovia xrj utio AEA eaxiv la/]. 8id xd auxd 8f) 
xal exdaxr) x£5v utio ABr, BrA, TAE ycovicov exaxepa xSv 
utco BAE, AEA eaxiv lory iaoycoviov dpa eaxl xo ABrAE 
Tievxdywvov. eSeiy^r) Se xal iaonXeupov. 

Eic; dpa xov Bcdevxa xuxXov nevxayiovov iaonXeupov xe 
xal iaoycbviov eyyeypanxai- onep e8ei noifjaai. 



IP'. 

nepi xov 8oi9evxa xuxXov Tievxdycovov iaoTcXeupov xe 
xai iaoycoviov Ttepiypdtjiai. 



in a given circle. 




Let ABCDE be the given circle. So it is required to 
inscribed an equilateral and equiangular pentagon in cir- 
cle ABCDE. 

Let the the isosceles triangle FGH be set up hav- 
ing each of the angles at G and H double the (angle) 
at F [Prop. 4.10]. And let triangle ACD, equiangular 
to FGH, have been inscribed in circle ABCDE, such 
that CAD is equal to the angle at F, and the (angles) 
at G and H (are) equal to ACD and CD A, respectively 
[Prop. 4.2]. Thus, ACD and CD A are each double 
CAD. So let ACD and CD A have been cut in half by 
the straight-lines CE and DB, respectively [Prop. 1.9]. 
And let AB, BC, DE and EA have been joined. 

Therefore, since angles ACD and CD A are each dou- 
ble CAD, and are cut in half by the straight-lines CE and 
DB, the five angles DAC, ACE, ECD, CDB, and BDA 
are thus equal to one another. And equal angles stand 
upon equal circumferences [Prop. 3.26]. Thus, the five 
circumferences AB, BC, CD, DE, and EA are equal to 
one another [Prop. 3.29]. Thus, the pentagon ABCDE 
is equilateral. So I say that (it is) also equiangular. For 
since the circumference AB is equal to the circumfer- 
ence DE, let BCD have been added to both. Thus, the 
whole circumference ABCD is equal to the whole cir- 
cumference EDCB. And the angle AED stands upon 
circumference ABCD, and angle BAE upon circumfer- 
ence EDCB. Thus, angle BAE is also equal to AED 
[Prop. 3.27]. So, for the same (reasons), each of the an- 
gles ABC, BCD, and CDE is also equal to each of BAE 
and AED. Thus, pentagon ABCDE is equiangular. And 
it was also shown (to be) equilateral. 

Thus, an equilateral and equiangular pentagon has 
been inscribed in the given circle. (Which is) the very 
thing it was required to do. 

Proposition 12 

To circumscribe an equilateral and equiangular pen- 
tagon about a given circle. 



120 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 




K r A 

"Eax« 6 Bodeic xuxXoc 6 ABrAE- 8eT Be Tiepi xov 
ABrAE xuxXov Tievxdyovov EaoTiXeupov xe xal iaoyoviov 
Tiepiypdijiai. 

Nevorfo'dti) xoO eyyeypa^evou Tievxaywvou xwv ytoviwv 
ar)^eTa xd A, B, T, A, E, coaxe lane, elvai xdc AB, Br, 
TA, AE, EA Tiepicpepeiac- xal Sid xwv A, B, V, A, E 
f]X"Q(^oay xoO xuxXou ecpoaixo^ievoa ai H0, 9K, KA, AM, 
MH, xal eiXr]tp , dw xou ABrAE xuxXou xevxpov xo Z, xal 
STieCeuxtiuaav ai ZB, ZK, ZT, ZA, ZA. 

Kal ETiel f] ^iev KA eu$eTa ecpdrcxexai xoO ABrAE xaxd 
xo T, arco Be xoO Z xevxpou era xf)v xaxd xo T eTiacpf)v 
CTieCeuxxai f) Zr, f] Zr dpa xdiJexoc eaxiv Ira xf)v KA- op-df) 
dpa eaxiv exaxepa xwv Tipoc iu T ytovicov. Bloc xd auxd 8f) 
xal ai Tipoc xolc B, A ar)fieioic ytoviai op-dai eiaiv. xal end 
opiDr] eaxiv f) utio ZrK y wvia, xo dpa arco xfjc ZK foov eaxi 
xolc aKO xov Zr, TK. 8id xd auxd 8f) xal xolc duo xwv ZB, 
BK i'aov laxl xo duo xfjc ZK- waxe xd aTio xwv Zr, TK 
xolc duo tuv ZB, BK eaxiv foa, Sv xo duo xfjc Zr iw dfto 
xfjc ZB eaxiv i'aov Xoitiov dpa xo diio xfjc TK xo diio xfjc 
BK eaxiv i'aov. for] dpa f] BK xfj TK. xal euel for] eaxlv 
f] ZB xrj Zr, xal xoivf] r] ZK, 8uo Bf] ai BZ, ZK Bual xdic 
rZ, ZK foai eiaiv xai pdaic f] BK pdaei xrj TK [eaxiv] fory 
ycovia dpa f] ^tev utio BZK [ycovia] xrj Otio KZT eaxiv for)- 
f] Be utio BKZ xrj utio ZKr- SmXfj dpa f] ^tev utio BZr xfjc 
utio KZr, f) 8e utio BKr xfjc utio ZKr. Bid xd auxd 8f] xai 
f] ^iev utio TZA xfjc utio TZA eaxi BuiXfj, fj 8e utio AAr 
xfjc utio ZAr. xal STid for] eaxlv f) Br Tiepicpepeia xrj TA, 
tar] eaxi xal ywvia f] utio BZr xrj utio TZA. xai eaxiv f) 
[Lev utio BZr xfjc utio KZr 8iTiXfj, f] Be utio AZr xfjc Otio 
AZr- for] dpa xal f) utio KZr xfj utio AZE eaxi 8e xal f] 
utio ZrK ytovia xfj utio ZrA i'ar). Suo Sf] xpiyovd eaxi xd 
ZKr, ZAr xdc Suo ywviac xdic Suai ywviaic foac exovxa 
xal h'iolv TiXeupdv uia TiXeupa iar]v xoivf]v auxGv xf]v ZE 
xal xdc XoiTidc dpa TiXeupdc xdic XoiTidic TiXeupdic i'aac ecei 
xal xf)v XoiTif]v ywviav xfj Xomfj ywvia- i'ar) dpa f] [lev KT 
eO'dsTa xfj TA, f) 8e utio ZKr ywvia xfj utio ZAr. xai etisI iar) 
eaxlv f) Kr xfj TA, SmXfj dpa f) KA xfjc Kr. Bid xd auxa Bf) 
8ei)(Tf)f]a£xai xal f) 8K xfjc BK BiTiXfj. xai eaxiv f) BK xfj Kr 
iar)- xal f) 9K dpa xfj KA eaxiv for). 6(ioitoc 8f) Beix'driaexai 



G 




K C L 



Let ABCDE be the given circle. So it is required 
to circumscribe an equilateral and equiangular pentagon 
about circle ABCDE. 

Let A, B, C, D, and E have been conceived as the an- 
gular points of a pentagon having been inscribed (in cir- 
cle ABCDE) [Prop. 3.11], such that the circumferences 
AB, BC, CD, DE, and EA are equal. And let GH, BK, 
KL, LM, and MG have been drawn through (points) A, 
B, C, D, and E (respectively), touching the circled And 
let the center F of the circle ABCDE have been found 
[Prop. 3.1]. And let FB, FK, FC, FL, and FD have 
been joined. 

And since the straight-line KL touches (circle) ABCDE 
at C, and FC has been joined from the center F to the 
point of contact C, FC is thus perpendicular to KL 
[Prop. 3.18]. Thus, each of the angles at C is a right- 
angle. So, for the same (reasons), the angles at B and 
D are also right-angles. And since angle FCK is a right- 
angle, the (square) on FK is thus equal to the (sum of 
the squares) on FC and CK [Prop. 1.47]. So, for the 
same (reasons), the (square) on FK is also equal to the 
(sum of the squares) on FB and BK. So that the (sum 
of the squares) on FC and CK is equal to the (sum of 
the squares) on FB and BK, of which the (square) on 
FC is equal to the (square) on FB. Thus, the remain- 
ing (square) on CK is equal to the remaining (square) 
on BK. Thus, BK (is) equal to CK. And since FB is 
equal to FC, and FK (is) common, the two (straight- 
lines) BF, FK are equal to the two (straight-lines) CF , 
FK. And the base BK [is] equal to the base CK. Thus, 
angle BFK is equal to [angle] KFC [Prop. 1.8]. And 
BKF (is equal) to FKC [Prop. 1.8]. Thus, BFC (is) 
double KFC, and BK C (is double) FKC. So, for the 
same (reasons), CFD is also double CFL, and DLC (is 
also double) FLC. And since circumference BC is equal 
to CD, angle BFC is also equal to CFD [Prop. 3.27]. 
And BFC is double KFC, and DFC (is double) LFC. 
Thus, KFC is also equal to LFC. And angle FCK is also 
equal to FCL. So, FKC and FLC are two triangles hav- 



121 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



xal exdaxr] xcov 0H, HM, MA exaxepa xcov 0K, KA icny 
laoTiXeupov apa eaxl to H0KAM Ttevxdycovov. Xeyco 5t|, 
oxi xal laoycoviov. enel yap for) eaxlv f\ Gtco ZKr ycovia xfj 
utio ZAr, xal eSeix^T) xfjc \ism Gtco ZKT BmXrj f\ utio 0KA, 
xfj? 8s utio ZAr 8mXrj f] utio KAM, xal f] 0tc6 0KA apa 
xfj utio KAM eaxiv Tar), o^oicoc 8f) Beiyjdriaexai xal exdaxr) 
xcov utio K6H, OHM, HMA exaxepa xcov uno 0KA, KAM 
larj- ai Tievxe apa ycovlai ai utio H0K, 6KA, KAM, AMH, 
MH0 laai aXXriXaic eiaiv. laoycoviov apa eaxl xo H9KAM 
Tievxdycovov. e8s[)(i9r) Se xal laoTcXEupov, xal Tcepiyeypanxai 
Tiepl xov ABrAE xuxXov. 

[Ilepl xov Scdevxa apa xuxXov TCEvxaycovov loonXeupov 
xe xal laoycoviov TtspiysypaTixai] • onep eBa Tioifjaai. 



ing two angles equal to two angles, and one side equal 
to one side, (namely) their common (side) FC. Thus, 
they will also have the remaining sides equal to the (cor- 
responding) remaining sides, and the remaining angle to 
the remaining angle [Prop. 1.26]. Thus, the straight-line 
KC (is) equal to CL, and the angle FKC to FLC. And 
since KC is equal to CL, KL (is) thus double KC. So, 
for the same (reasons), it can be shown that H K (is) also 
double BK. And BK is equal to KC. Thus, HK is also 
equal to KL. So, similarly, each of HG, CM, and ML 
can also be shown (to be) equal to each of HK and KL. 
Thus, pentagon GHKLM is equilateral. So I say that 
(it is) also equiangular. For since angle FKC is equal to 
FLC, and HKL was shown (to be) double FKC, and 
KLM double FLC, HKL is thus also equal to KLM. 
So, similarly, each of KHG, HGM, and GML can also 
be shown (to be) equal to each of HKL and KLM. Thus, 
the five angles GHK, HKL, KLM, LMG, and MGH 
are equal to one another. Thus, the pentagon GHKLM 
is equiangular. And it was also shown (to be) equilateral, 
and has been circumscribed about circle ABCDE. 

[Thus, an equilateral and equiangular pentagon has 
been circumscribed about the given circle]. (Which is) 
the very thing it was required to do. 



t See the footnote to Prop. 3.34. 



If. 

Fic, xo 8oil>£v Tcevxdyovov, 6 saxiv taoTiXeupov xe xal 
laoycoviov, xuxXov Eyypd^ai. 

A 




r k a 

"Eaxco xo BoiJev Ttevxdycovov laoTtXeupov xe xal laoycovi- 
ov xo ABrAE- 6el 6f] tic, xo ABrAE Tcevxdycovov xuxXov 
eyypdtjiai. 

Texuf]a , dco yap exaxepa xcov utio BIA, TAE ycovicov 
8[)(a utco exaxepa? xcov TZ, AZ eu'deicov xal aTco xou Z 
arjudou, xaO' o aujipdXXouaiv dXXrjXau; ai TZ, AZ euiMai, 
eTte^eu)cdcoaav ai ZB, ZA, ZE eui9elai. xal eTcel Tar) eaxlv 



Proposition 13 

To inscribe a circle in a given pentagon, which is equi- 
lateral and equiangular. 

A 




C K D 

Let ABCDE be the given equilateral and equiangular 
pentagon. So it is required to inscribe a circle in pentagon 
ABCDE. 

For let angles BCD and CDE have each been cut 
in half by each of the straight-lines CF and DF (re- 
spectively) [Prop. 1.9]. And from the point F, at which 
the straight-lines CF and DF meet one another, let the 



122 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



f) Br xfj TA, xoivf] Be f] TZ, 860 8f) ai Br, TZ Suol xaic; 
Ar, rZ iaai eiaiv xal yiovia f) (mo BrZ ycovia tfi bub 
ArZ [eaxiv] for) - pdaic; dpa f] BZ pdaei xfj AZ eaxiv iar], 
xal xo BrZ xpiywvov xw ArZ xpiywvw eaxiv laov, xal ai 
Xoinai ycoviai xaic; XomaTc; ycoviaic; laai eaovxai, ucp' ac al 
laai rcXeupal ujraxeivouaiv for) apa f] (mo TBZ ycovia xfj 
O716 TAZ. xal en:el BiuXfj eaxiv f] Otco IAE xfjc; bub IAZ, 
Xat] 8e f] (jlsv Otto TAE xfj Otto ABr, f] Se uko TAZ xfj hub 
TBZ, xal f] uuo TBA dpa xfjc; utco TBZ eaxi SiitXfj- for] 
apa f] utco ABZ ycovia xfj \mb ZBE f] apa Otto ABr ycovia 
8[)(a xex^rjxai Otto xfjc; BZ sbftemc,. o^toicoc; Bf] 8ei)(iL>r)aexai, 
oxi xal exaxepa xcov bub BAE, AEA Bi)(a xex^rjxai bub 
exaxepac; xcov ZA, ZE e0i!)eic5v. fj)(Tf)waav 8f] duo xou Z 
ar^eiou etxI xac; AB, Br, TA, AE, EA eOiJeiac; xd-dexoi 
ai ZH, Z0, ZK, ZA, ZM. xal creel tar] eaxiv f) Otio 9rZ 
ycovia xfj bnb KFZ, eaxi 8e xal opiDf] f) Oreo Z8r [6pi!)fj] 
xfj Oreo ZKr for), 860 8f] xpiycovd eaxi xd Z6r, ZKr xac; 
860 ycoviac; 8ual ycoviaic; laac; exovxa xa ' t^ av TtXeupdv 
TcXeupa iar)v xoivfjv auxcov xf)v Zr Orcoxeivouaav bnb [iiav 
xcov lacov ycovicov xal xac; XoiTtdc; apa TtXeupac; xaic; XoiTcdic; 
TiXeupdic; laac; '€E,zv Tar) apa f) Z9 xdiitexoc; xf) ZK xai9exco. 
ojioicoc; 8f) SeivOfjaexai, oxi xal exdaxr) xcov ZA, ZM, ZH 
exaxepa xcov Z0, ZK for] eaxiv ai rcevxe dpa eO-delai ai ZH, 
Z6, ZK, ZA, ZM foai dXXfjXaic; eiaiv. 6 apa xevxpco xco Z 
8iaaxf][iaxi 8e evi xcov H, 0, K, A, M xuxXoc; ypacpojievoc; 
fjc;ei xal 81a xwv Xoitccov arj^ieicov xal ecpdij^exai xwv AB, Br, 
TA, AE, EA e0i5eicov 81a xo opiSdc; eTvai xac; rcpoc; xou; H, 
0, K, A, M ar^eioic; ycoviac;. si yap oOx e(pd<}>exai aOxcov, 
dXXd xe^iel auxdc;, aujj.pf]aexai xf]v xfj 8ia[iexpco xou xuxXou 
npoc; op-ddc; die' axpac; dyojievrjv evxoc; TUTtxeiv xou xuxXou' 
oTiep axoKov eBeivdr). oux apa 6 xsvxpw xfi Z 8iaaxf]^.axi 8e 
evi xcov H, &, K, A, M arjpieiwv ypacpo^tevoc; xuxXoc; xe^tel 
xac; AB, Br, EA, AE, EA eb-deiac,- ecpdi^exai apa auxfiv. 
yeypdcpi^w wc; 6 H0KAM. 

Eic; apa xo So-dev Tievxdyovov, 6 eaxiv iaoiiXeupov xe 
xal iaoyoviov, xuxXoc; eyyeypanxai- onep e8ei uoifjaai. 



i5'. 

nepl xo 8oiL>ev iievxdywvov, o eaxiv iaonXeupov xe xal 
iaoycoviov, xuxXov ixepiypd^ai. 

^axo xo 8oT5ev uevxdywvov, 6 eaxiv iaoTiXeupov xe xal 



straight-lines FB, FA, and FE have been joined. And 
since BC is equal to CD, and CF (is) common, the two 
(straight-lines) BC, CF are equal to the two (straight- 
lines) DC, CF. And angle BCF [is] equal to angle 
DCF. Thus, the base BF is equal to the base DF, and 
triangle BCF is equal to triangle DCF, and the remain- 
ing angles will be equal to the (corresponding) remain- 
ing angles which the equal sides subtend [Prop. 1.4]. 
Thus, angle CBF (is) equal to CDF. And since CDF 
is double CDF, and CDE (is) equal to ABC, and CDF 
to CBF, CBA is thus also double CBF. Thus, angle 
ABF is equal to F BC. Thus, angle ABC has been cut 
in half by the straight-line BF. So, similarly, it can be 
shown that BAE and AED have been cut in half by the 
straight-lines FA and FE, respectively. So let FG, FH, 
FK, FL, and FM have been drawn from point F, per- 
pendicular to the straight-lines AB, BC, CD, DE, and 
EA (respectively) [Prop. 1.12]. And since angle HCF 
is equal to KCF, and the right-angle FHC is also equal 
to the [right-angle] FKC, FHC and FKC are two tri- 
angles having two angles equal to two angles, and one 
side equal to one side, (namely) their common (side) FC, 
subtending one of the equal angles. Thus, they will also 
have the remaining sides equal to the (corresponding) 
remaining sides [Prop. 1.26]. Thus, the perpendicular 
FH (is) equal to the perpendicular FK. So, similarly, it 
can be shown that FL, FM, and FG are each equal to 
each of FH and FK. Thus, the five straight-lines FG, 
FH, FK, FL, and FM are equal to one another. Thus, 
the circle drawn with center F, and radius one of G, H, 
K, L, or M, will also go through the remaining points, 
and will touch the straight-lines AB, BC, CD, DE, and 
EA, on account of the angles at points G, H, K, L, and 
M being right-angles. For if it does not touch them, but 
cuts them, it follows that a (straight-line) drawn at right- 
angles to the diameter of the circle, from its extremity, 
falls inside the circle. The very thing was shown (to be) 
absurd [Prop. 3.16]. Thus, the circle drawn with center 
F, and radius one of G, H, K, L, or M, does not cut 
the straight-lines AB, BC, CD, DE, or EA. Thus, it will 
touch them. Let it have been drawn, like GHKLM (in 
the figure). 

Thus, a circle has been inscribed in the given pen- 
tagon which is equilateral and equiangular. (Which is) 
the very thing it was required to do. 

Proposition 14 

To circumscribe a circle about a given pentagon which 
is equilateral and equiangular. 

Let ABCDE be the given pentagon which is equilat- 



123 



ETOIXEIftN 5'. 



ELEMENTS BOOK 4 



iaoycoviov, to ABEAE- 8sT 8f) itepi to ABrAE TtsvTdycovov 
xuxXov Tiepiypd(J>ai. 

A 




TeT[if]a , dco §r) sxaTspa twv Otto BrA, EAE yioviSSv 
Sixa Otto sxaT£pa<; t£>v TZ, AZ, xod arco tou Z ar)(jisiou, 
xa-d' o aujipdXXouoiv ai Eui&Elai, Em. Ta B, A, E ar]\isia 
£Tie^£U)fd«aav sMsTai ai ZB, ZA, ZE. ojioiwc; 8/) tG Ttpo 
toutou 6ei)cdr]0£Tai, oti xod sxaaTT) twv Otto TBA, BAE, 
AEA yiovifiv 8[)(a T£T|ir]Tai Otto exdoTT)<; tGv ZB, ZA, ZE 
eut}eiGv. xod ettel Tor) eotIv r) Otto BTA ycovia Tfj Otto EAE, 
xai egti Tfj<; ^iev Otto BrA f][iia£ia r) Otto ZEA, Tfjg 8s Otto 
EAE f^iasia f] Otto EAZ, xod r] Otto ZrA dpa Tfj Otto ZAr 
egtiv icy]- Sots xod nXsupd f) ZT TiXsupa Tfj ZA eotlv lor). 
by.o'\.oic, 8r) 8ei)cdr]0£Tai, oti xod exdoTT) tGv ZB, ZA, ZE 
sxaTspa t«v Zr, ZA saw tar)- ai tievts dpa su-daai ai 
ZA, ZB, Zr, ZA, ZE i'oai dXXfjXaig siaiv. 6 dpa xsvTpcp 
tw Z xai 8iaaTr]jj.aTi svi t£Sv ZA, ZB, ZT, ZA, ZE xuxXoc; 
ypacpojisvoc; ffesi. xal 8id t«v Xoitmov a/jjisiwv xod eaxoti tee- 
piysypajijjisvoc;. TtEpiysypdcp^co x °d eo"tm 6 ABrAE. 

ILepi dpa to 8ot5ev TiEVTdyojvov, 6 eotiv iaoTtXsupov te 
xal iaoycoviov, xuxXoc TCEpiysypaTtTai' ousp eSel Tioirjaai. 



is'. 

El? tov 8oiD£VTa xuxXov s^dyiovov iaonXsupov te xal 
iaoycoviov Eyypdtjiai. 

'EaTCO 6 So-dslc; xuxXo<; 6 ABEAEZ- SeT Sf] sic; tov 
ABrAEZ xuxXov sc;dycovov iaoTtXsupov te xai iaoycoviov 
Eyypdtjjai. 

"H)(Tf)w toO ABrAEZ xuxXou 8id^iSTpo<; f) AA, xai 
siXficp'dco to XEVTpov toO xuxXou to H, xai xsvTpcp jisv 
tw A 8iaaTT]^aTi 8s iu AH xuxXoc; ysypdcpiJco 6 EHr0, 
xai £7u£EU)cdsTaai ai EH, TH Sifix^coaav ski Ta B, Z ar)[i£la, 
xai ETCEc'Eu/'dcoaav ai AB, Br, TA, AE, EZ, ZA- Xsyco, oti 



eral and equiangular. So it is required to circumscribe a 
circle about the pentagon ABCDE. 

A 




So let angles BCD and CDE have been cut in half by 
the (straight-lines) CF and DF, respectively [Prop. 1.9]. 
And let the straight-lines FB, FA, and FE have been 
joined from point F, at which the straight-lines meet, 
to the points B, A, and E (respectively). So, similarly, 
to the (proposition) before this (one), it can be shown 
that angles CBA, BAE, and AED have also been cut 
in half by the straight-lines FB, FA, and FE, respec- 
tively. And since angle BCD is equal to CDE, and FCD 
is half of BCD, and CDF half of CDE, FCD is thus 
also equal to FDC. So that side FC is also equal to side 
FD [Prop. 1.6]. So, similarly, it can be shown that FB, 
FA, and FE are also each equal to each of FC and FD. 
Thus, the five straight-lines FA, FB, FC, FD, and FE 
are equal to one another. Thus, the circle drawn with 
center F, and radius one of FA, FB, FC, FD, or FE, 
will also go through the remaining points, and will have 
been circumscribed. Let it have been (so) circumscribed, 
and let it be ABCDE. 

Thus, a circle has been circumscribed about the given 
pentagon, which is equilateral and equiangular. (Which 
is) the very thing it was required to do. 

Proposition 15 

To inscribe an equilateral and equiangular hexagon in 
a given circle. 

Let ABCDEF be the given circle. So it is required to 
inscribe an equilateral and equiangular hexagon in circle 

ABCDEF. 

Let the diameter AD of circle ABCDEF have been 
drawn, t and let the center G of the circle have been 
found [Prop. 3.1]. And let the circle EGCH have been 
drawn, with center D, and radius DC And EG and CG 
being joined, let them have been drawn across (the cir- 



124 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



to ABrAEZ s^aytovov iaonXeupov xe eaxi xal iaoyioviov. 



e 




A 



Tkei yap xo H ar^elov xevxpov sail xou ABrAEZ 
xuxXou, igt] eaxlv f) HE xrj HA. raiXiv, end to A a/jjiaov 
xevxpov sail toO Hr0 xuxXou, for) eaxlv r) AE xrj AH. 
dXX' f\ HE xrj HA eSelx'dr) tar) - xal f] HE apa xrj EA tar) 
eaxlv iaoTtXeupov apa eaxi xo EHA xplycovov xal ai xpeu; 
apa auxou yoviai ai bnb EHA, HAE, AEH laai dXXr]Xai<; 
eiaiv, C7iei8r]TT;ep xwv EaoaxsXwv xpiycivov ai Kpo<; xrj pdaei 
ywviai laai dXXrjXaic eialv xal eiaiv ai xpeu; xou xpiywvou 
ywviai Buaiv op^aTc; low f] apa urab EHA ywvla xplxov eaxl 
860 6pif)Gv. 6^iolo<; 5r] Seix'driaexai xal f\ bnb AHT xplxov 
860 op^wv. xal ckcI f] TH eu'deTa em xf]v EB axa'delaa xa<; 
ecpe£rj<; ywvla<; xa<; bnb EHr, THB Sualv opiJau; laac noieT, 
xal Xomrj apa f] O716 THB xplxov eaxl 860 op-dGv ai apa 
bnb EHA, AHr, THB ywvlai laai dXXr]Xai<; eialv oaxe xal 
ai xaxa xopucprjv auxdic ai bnb BHA, AHZ, ZHE i'aai eialv 
[xdlc bnb EHA, AHr, THB]. ai e? apa ywvlai ai (mo EHA, 
AHr, THB, BHA, AHZ, ZHE I'aai dXX^Xai.; eiaiv. ai 8e 
laai ywvlai era lawv Ttepicpepeiwv peprjxaaiv al ec; apa Tte- 
picpepeiai ai AB, Br, EA, AE, EZ, ZA i'aai dXXr]Xaic eialv. 
bnb 8e xa<; laac; Ttepicpepelac al laai eu-deTai UTtoxeivouaiv 
ai §5 apa eu'delai laai dXXf|Xai<; eiaiv iaorcXeupov apa eaxl 
xo ABrAEZ eQaywvov. Xeyw §7), 6x1 xal iaoywviov. emci 
yap lot] eaxlv f] ZA Ttepicpepeia xrj EA Ttepicpepeia, xoivr] 
TipoaxelaiDw f] ABrA Ttepicpepeia- oXr] apa f) ZABrA oXr] 
xrj EArBA eaxiv I'ar)- xal peprjxev era ^tev xfj? ZABrA 
Ttepicpepeia<; f] bnb ZEA ywvla, era 8e xfj? EArBA Ttepi- 
cpepelac; rj bnb AZE yovia - iar] apa f) tmo AZE ywvia xfj 
bnb AEZ. o^iolcoc; 8r) 8ei)edr|aexai, 6x1 xal ai XoiTtai ywvlai 
xou ABrAEZ e^aywvou xaxa piiav laai eiaiv exaxepa xwv 
bnb AZE, ZEA ywviwv iaoywviov apa eaxl xo ABrAEZ 
e^dywvov. eSeix^r] 8e xal iaoTtXeupov xal eyyeypauxai eic; 
xov ABrAEZ xuxXov. 

Eiz apa xov BoiJevxa xuxXov e^dywvov iaoTtXeupov xe 



cle) to points B and F (respectively). And let AB, BC, 
CD, BE, EF, and FA have been joined. I say that the 
hexagon ABCDEF is equilateral and equiangular. 



H 




A 



For since point G is the center of circle ABCDEF, 
GE is equal to CD. Again, since point D is the cen- 
ter of circle GCH, DE is equal to DC. But, GE was 
shown (to be) equal to GD. Thus, GE is also equal to 
ED. Thus, triangle EGD is equilateral. Thus, its three 
angles EGD, GDE, and DEG are also equal to one an- 
other, inasmuch as the angles at the base of isosceles tri- 
angles are equal to one another [Prop. 1.5]. And the 
three angles of the triangle are equal to two right-angles 
[Prop. 1.32]. Thus, angle EGD is one third of two right- 
angles. So, similarly, DGC can also be shown (to be) 
one third of two right-angles. And since the straight-line 
CG, standing on EB, makes adjacent angles EGG and 
CGB equal to two right-angles [Prop. 1.13], the remain- 
ing angle CGB is thus also one third of two right-angles. 
Thus, angles EGD, DGC, and CGB are equal to one an- 
other. And hence the (angles) opposite to them BGA, 
AGF, and FGE are also equal [to EGD, DGC, and 
CGB (respectively)] [Prop. 1.15]. Thus, the six angles 
EGD, DGC, CGB, BGA, AGF, and FGE are equal 
to one another. And equal angles stand on equal cir- 
cumferences [Prop. 3.26]. Thus, the six circumferences 
AB, BC, CD, DE, EF, and FA are equal to one an- 
other. And equal circumferences are subtended by equal 
straight-lines [Prop. 3.29]. Thus, the six straight-lines 
{AB, BC, CD, DE, EF, and FA) are equal to one 
another. Thus, hexagon ABCDEF is equilateral. So, 
I say that (it is) also equiangular. For since circumfer- 
ence FA is equal to circumference ED, let circumference 
ABCD have been added to both. Thus, the whole of 
FABCD is equal to the whole of EDCBA. And angle 
FED stands on circumference FABCD, and angle AFE 
on circumference EDCBA. Thus, angle AFE is equal 



125 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



xal iaoywviov EYY^YP a7txal ' OTI£ P e8ei noifjaai. 



I16pia[Jia. 

'Ex 5r) xouxou (pavepov, oxi f) xoO e^aywvou TtXeupa tar) 
saxl xrj ex xoO xevxpou xoO xuxXou. 

Ojioicdc; 8e xou; era tou Tievxaycbvou eav 8ia xcov xaxa 
xov xuxXov Siaipeaewv ecpaTtxojievac; xou xuxXou aYayt^ev, 
TxspiYpacpr^aexai Ttepl xov xuxXov e^aYWvov laoTtXeupov 
xe xal iooywviov dxoXoui&Mt; xou; ski xoO TievxaY^vou 
eipr^evoig. xal exi Bia xwv o^ioicov xou; etu xou nevxaYCovou 
eipr^evoi? sic, xo Bcdsv s^otYCOvov xuxXov eYyp6n\>o[Lev t£ 
xal TiepiYpa^o^tEv oTiep e5ei Ttoirjaai. 

t See the footnote to Prop. 4.6. 

El? xov ScfJevxa xuxXov TisvxexaiBexaYCOvov laonXsupov 
xe xal EaoYCoviov tyYpcn\)ai. 

A 




Tiaxo 6 Bo-deu; xuxXo? 6 ABIA- SeT Br] sic, xov ABrA 
xuxXov nevxExaiSexaYCovov laoitXeupov xe xal Iooycoviov 

'EyysyP^V'S" £ k TOV ABrA xuxXov xpiY«vou \iev tao- 
TiXeupou xoO zlc, auxov eYYP°«po^evou TtXeupa f) Ar, Ttev- 
xaYWvou 8s laoiiXeupou f\ AB - oiwv apa eaxlv 6 ABrA 
xuxXoc Tacov xp^axwv Bexaravxe, xoiouxmv f] [ie\ ABT 
Txepicpepeia xptxov ouaa xou xuxXou eaxai Ttevxe, f) Be AB 
Txepicpepeia Tte^txov ouaa xou xuxXou eaxai xpiwv Xomf] apa 



to DEF [Prop. 3.27]. Similarly, it can also be shown 
that the remaining angles of hexagon ABC DEF are in- 
dividually equal to each of the angles AFE and FED. 
Thus, hexagon ABC DEF is equiangular. And it was also 
shown (to be) equilateral. And it has been inscribed in 
circle ABCDE. 

Thus, an equilateral and equiangular hexagon has 
been inscribed in the given circle. (Which is) the very 
thing it was required to do. 

Corollary 

So, from this, (it is) manifest that a side of the 
hexagon is equal to the radius of the circle. 

And similarly to a pentagon, if we draw tangents 
to the circle through the (sixfold) divisions of the (cir- 
cumference of the) circle, an equilateral and equiangu- 
lar hexagon can be circumscribed about the circle, analo- 
gously to the aforementioned pentagon. And, further, by 
(means) similar to the aforementioned pentagon, we can 
inscribe and circumscribe a circle in (and about) a given 
hexagon. (Which is) the very thing it was required to do. 



Proposition 16 

To inscribe an equilateral and equiangular fifteen- 
sided figure in a given circle. 

A 




Let ABCD be the given circle. So it is required to in- 
scribe an equilateral and equiangular fifteen-sided figure 
in circle ABCD. 

Let the side AC of an equilateral triangle inscribed 
in (the circle) [Prop. 4.2], and (the side) AB of an (in- 
scribed) equilateral pentagon [Prop. 4.11], have been in- 
scribed in circle ABCD. Thus, just as the circle ABCD 
is (made up) of fifteen equal pieces, the circumference 
ABC, being a third of the circle, will be (made up) of five 



126 



ETOIXEIfiN 5'. 



ELEMENTS BOOK 4 



f] Br xov Tacov 860. xexjnqcrdGj f\ Br Sixa xaxa to E- 
sxaxspa apa xfiiv BE, Er TCpicpspeifiv ravxexaiSexaxov eaxi 
xou ABrA xuxXou. 

'Eav apa £Tu£euc;avxec; Tag BE, Er I'aac; auxau; xaxa to 
auvsy^ su-deiac; svap^oawjisv sic; xov ABrA[E] xuxXov, 
eaxai tic, auxov £yy£YP°W£ vov TievxExaiBexdycovov iaorcXeu- 
pov xe xal laoyoviov ojtep eSei Ttoirjaai.. 

'Ojioiwc; 8e xou; era xou nevxaYWvou eav 81a xfiv 
xaxa xov xuxXov 8iaipeaewv ecpauxo^svac; xou xuxXou 
dyaYW^ev, TtepiYpacprjaexai Ttepl xov xuxXov Ttevxexai- 
8exaYWvov laoitXeupov xe xal iooywviov. erxi 8e Sid 
xwv ojioiwv xolc; era xou nsvxaYWvou Seli;e«v xal etc; xo 
Bo'dev TievxexaiBexaYWvov xuxXov eyy p6n\to[ie\i T £ xal ne- 
piYpdtjjo^iev oiiep eSei Ttoirjaai.. 



such (pieces), and the circumference AB, being a fifth of 
the circle, will be (made up) of three. Thus, the remain- 
der BC (will be made up) of two equal (pieces). Let (cir- 
cumference) BC have been cut in half at E [Prop. 3.30]. 
Thus, each of the circumferences BE and EC is one fif- 
teenth of the circle ABODE. 

Thus, if, joining BE and EC, we continuously in- 
sert straight-lines equal to them into circle ABCD[E] 
[Prop. 4.1], then an equilateral and equiangular fifteen- 
sided figure will have been inserted into (the circle). 
(Which is) the very thing it was required to do. 

And similarly to the pentagon, if we draw tangents to 
the circle through the (fifteenfold) divisions of the (cir- 
cumference of the) circle, we can circumscribe an equilat- 
eral and equiangular fifteen-sided figure about the circle. 
And, further, through similar proofs to the pentagon, we 
can also inscribe and circumscribe a circle in (and about) 
a given fifteen-sided figure. (Which is) the very thing it 
was required to do. 



127 



128 



ELEMENTS BOOK 5 

Proportion 



tThe theory of proportion set out in this book is generally attributed to Eudoxus of Cnidus. The novel feature of this theory is its ability to deal 
with irrational magnitudes, which had hitherto been a major stumbling block for Greek mathematicians. Throughout the footnotes in this book, 
a, ft, 7, etc., denote general (possibly irrational) magnitudes, whereas m, n, I, etc., denote positive integers. 



129 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



"Opoi. 

a'. Mepog taxi (ieyeiL>oc; \Leye r &o\jQ xo IXaaoov xou 
[idZovoz, oxav xaxajiexpfj to \is%o\. 

P'. IToXXaTtXaoiov 8s to jieT^ov xou eXdxxovoc, oxav xa- 
xatiexprjxai Otto xou eXdxxovoc;. 

y'. Aoyog eoxl 860 y.sys-&&\i ojioyevSiv f] xaxd tltj- 
Xixox/jxd n:oia a/eaic;. 

8'. Aoyov l)(£iv Tipoc; aXXr)Xa (ieye'dr) Xeyexai, a Suvaxai 
TtoXXaTiXaoia^ojjeva dXXf|Xiov UTtepe/eiv. 

e'. 'Ev xw auxC) Xoyw jieye'dr] Xeyexai etvai TtpGxov 
Tipoc; 8euxepov xal xpixov Tipoc; xexapxov, oxav xd xou 
Tipcixou xa[ xpixou lodxic; TioXXaTiXdoia x«v xou 8euxepou 
xal xexdpxou iodxic; KoXXanXaaicov xaif)' otioiovouv TioXXa- 
TiXaoiaojiov exdxepov exaxepou fj ajja bnzpixT] °V a l ' CTa fl 
fj djjia eXXeiTifj Xrjcpdevxa xaxdXX/jXa. 

9'. Td Se xov auxov e/ovxa Xoyov (jieyeiSr) dvdXoyov 
xaXeurdco. 

C. "Oxav Se xfiv iodxic; TioXXaTiXaoitov xo jjcv xou 
Tipcoxou TtoXXanXdoiov UTiepexn TO ° TO ° Beuxepou tioX- 
XaTiXaoiou, xo 8e xou xpixou TioXXaTiXdaiov [ir\ UTiepexT) 
xou xou xexdpxou TioXXaTiXaaiou, xoxe xo npwxov npbc, xo 
Beuxepov jjeii^ova Xoyov e^eiv Xeyexai, fjTiep xo xpixov Tipoc; 
xo xexapxov. 

/)'. AvaXoyia 8e ev xpiolv opoig ekayioTf] eoxiv. 

•&'. "Oxav Be xpia jieye'dr] dvdXoyov fj, xo Tipoxov Tipoc 
xo xpixov BmXaoiova Xoyov e/eiv Xeyexai f)7tep Tipoc; xo 
Seuxepov. 

1'. "Oxav 8e xeooapa (icye^T) dvdXoyov fj, xo Ttpcoxov 
Tipoc; xo xexapxov xpmXaoiova Xoyov e)(eiv Xeyexai fjTiep 
Tipoc; xo Beuxepov, xal del tEjff, ojjlolcoc;, 6? av f] dvaXoyia 
UTidpxr). 

ia'. 'OjioXoya (leyeiSr] Xeyexai xd jjcv f]youjieva xou; 
f]you^evoi<; xd 8e CTtojjeva xolc; enojievoic. 

iP'. 'EvaXXd^ Xoyog eoxl Xfjcjxc; xou fjyoujjevou Tipoc; xo 
fjyoujievov xal xou ercojjievou Tipoc; xo eixojievov. 

iy'. AvaTiaXiv Xoyog eoxl Xfjcjiu; xou enojievou tbc; 
fjyou^evou Tipoc; xo fjyoufjievov cbc; ctiojjcvov. 

18'. Euvdeoic; Xoyou eoxl Xrjcjnc; xou fjyou^ievou (iexd xou 
enojievou (be evoc Tipoc; auxo xo euojievov. 

ie'. Aiaipeou; Xoyou eoxl Xfjcjiic xfjc UTiepoxfjc;, fj unepe/ei 
xo f)you(ievov xou enojievou, Tipoc; auxo xo e7i6[ievov. 

19'. 'Avaoxpocpr) Xoyou eoxl Xfj(J>ic; xou f)you(ievou npoc 
xrjv U7iepo);r]v, f) unepe/ei xo fjyoujievov xou C7io|jievou. 

iC- Ai° ioou Xoyoc; eoxl nXeiovwv ovxtov [ieycdGv xal 
dXXwv auxolg Tocov xo nX^oc ouvSuo Xajj.pavojj.evwv xal 
ev x£S auxw Xoyco, oxav fj &>c, ev xolc; upcoxoig jjeye-deoi xo 
npSxov Tipoc; xo eo^axov, ouxwc; ev xolg Seuxepoic; jjeye-deoi 
xo npSxov Tipog xo eo/axov fj aXXwg- Xfj^it; x£5v axpcov 



Definitions 

1. A magnitude is a part of a(nother) magnitude, the 
lesser of the greater, when it measures the greater.^ 

2. And the greater (magnitude is) a multiple of the 
lesser when it is measured by the lesser. 

3. A ratio is a certain type of condition with respect to 
size of two magnitudes of the same kind.-f 

4. (Those) magnitudes are said to have a ratio with re- 
spect to one another which, being multiplied, are capable 
of exceeding one another. § 

5. Magnitudes are said to be in the same ratio, the first 
to the second, and the third to the fourth, when equal 
multiples of the first and the third either both exceed, are 
both equal to, or are both less than, equal multiples of the 
second and the fourth, respectively, being taken in corre- 
sponding order, according to any kind of multiplication 
whatever. 1 

6. And let magnitudes having the same ratio be called 
proportional.* 

7. And when for equal multiples (as in Def. 5), the 
multiple of the first (magnitude) exceeds the multiple of 
the second, and the multiple of the third (magnitude) 
does not exceed the multiple of the fourth, then the first 
(magnitude) is said to have a greater ratio to the second 
than the third (magnitude has) to the fourth. 

8. And a proportion in three terms is the smallest 
(possible). 8 

9. And when three magnitudes are proportional, the 
first is said to have to the third the squaredll ra tio of that 
(it has) to the second.^ 

10. And when four magnitudes are (continuously) 
proportional, the first is said to have to the fourth the 
cubedW ratio of that (it has) to the second. §§ And so on, 
similarly, in successive order, whatever the (continuous) 
proportion might be. 

11. These magnitudes are said to be corresponding 
(magnitudes): the leading to the leading (of two ratios), 
and the following to the following. 

12. An alternate ratio is a taking of the (ratio of the) 
leading (magnitude) to the leading (of two equal ratios), 
and (setting it equal to) the (ratio of the) following (mag- 
nitude) to the following. ^ 

13. An inverse ratio is a taking of the (ratio of the) fol- 
lowing (magnitude) as the leading and the leading (mag- 
nitude) as the following.** 

14. A composition of a ratio is a taking of the (ratio of 
the) leading plus the following (magnitudes), as one, to 
the following (magnitude) by itself. $$ 



130 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



xocd' UKS^aipeoiv xwv (irotov. 15. A separation of a ratio is a taking of the (ratio 

it)'. Tsxapaypievr] 8e dvaXoyia eoxiv, oxav xpiGv ovxwv of the) excess by which the leading (magnitude) exceeds 
^eyei^Gv xod dXXwv auxolc; icxov xo nXfj'doc yivrjxai <i>c; [iev the following to the following (magnitude) by itself. I 
sv xoic; Tipcoxoic; [Leye-Qeoiv f]youjj.£vov Ttpoc; sttojisvov, ouxgjc 16. A conversion of a ratio is a taking of the (ratio 
sv xoic; Beuxepou; ^Leyen&Eaiv f]you^i£vov Tipoc; etto^evov, tbg of the) leading (magnitude) to the excess by which the 
8e ev ioXc, upcoxoic; jisys-decnv etto^evov npbc, dXXo xi, ouxwg leading (magnitude) exceeds the following. W 
sv xolc; Seuxepou; dXXo xi Tipoc rjyou^evov. 17. There being several magnitudes, and other (mag- 

nitudes) of equal number to them, (which are) also in the 
same ratio taken two by two, a ratio via equality (or ex 
aequali) occurs when as the first is to the last in the first 
(set of) magnitudes, so the first (is) to the last in the sec- 
ond (set of) magnitudes. Or alternately, (it is) a taking of 
the (ratio of the) outer (magnitudes) by the removal of 
the inner (magnitudes). 

18. There being three magnitudes, and other (magni- 
tudes) of equal number to them, a perturbed proportion 
occurs when as the leading is to the following in the first 
(set of) magnitudes, so the leading (is) to the following 
in the second (set of) magnitudes, and as the following 
(is) to some other (i.e., the remaining magnitude) in the 
first (set of) magnitudes, so some other (is) to the leading 
in the second (set of) magnitudes. §§§ 

t In other words, a is said to be a part of (3 if f3 = m a. 

t In modern notation, the ratio of two magnitudes, a and (3, is denoted a : (3. 

§ In other words, a has a ratio with respect to f3 if m a > f3 and n/3 > a, for some m and n. 

' In other words, a : /3 : : 7 : <5 if and only if m a > n/3 whenever m 7 > n S, and ma = n/3 whenever m 7 = n 8, and ma < n/3 whenever 
rn 7 < n 8, for all m and n. This definition is the kernel of Eudoxus' theory of proportion, and is valid even if a, (3, etc., are irrational. 
* Thus if a and (3 have the same ratio as 7 and <5 then they are proportional. In modern notation, a : (3 :: 7 : <5. 
$ In modern notation, a proportion in three terms — a, f3, and 7 — is written: a : f3 :: f3 : 7. 
II Literally, "double". 

tt In other words, if a : (3 :: (3 : 7 then a : 7 :: a 2 : (3 2 . 
H Literally, "triple". 

In other words, if a : (3 :: f3 : 7 7 : <5 then a : 8 :: a 3 : f3 3 . 
■ ■ In other words, if a : (3 :: 7 : <5 then the alternate ratio corresponds to a : 7 :: (3 : 8. 
** In other words, if a : (3 then the inverse ratio corresponds to f3 : a. 
$$ In other words, if a : (3 then the composed ratio corresponds to a + (3 : f3. 
II II In other words, if a : (3 then the separated ratio corresponds to a — f3 : f3. 
ttt In other words, if a : (3 then the converted ratio corresponds to a : a — (3. 

it* In other words, if a, (3, 7 are the first set of magnitudes, and 5, e, ( the second set, and a : f3 : 7 :: 8 : e : then the ratio via equality (or ex 
aequali) corresponds to a : 7 :: <5 : ( . 

§§§ In other words, if a, (3, 7 are the first set of magnitudes, and <5, e, ( the second set, and a : (3 :: 8 : e as well as (3 : 7 :: ( : 8, then the proportion 
is said to be perturbed. 

a'. Proposition V 

'Eav rj onoaaouv jisys-d/) otcogcovouv jieye^wv Tacov xo If there are any number of magnitudes whatsoever 
TzkyjOoQ excxgxov ex&oxou iadxi<; ttoXXootMchov, oacmXdaiov (which are) equal multiples, respectively, of some (other) 
saxiv ev x«v ^sys'dcSv smoq, xooauxaTiXdoia eaxou xal xd magnitudes, of equal number (to them), then as many 



131 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



A H B T A 

i 1 1 i 1 1 

Ei 1 Z' 1 

TSaxw OTOaaouv ^ey^dr] xd AB, FA otcogcovoOv [ls- 
ys-dcov twv E, Z Tacov to TtXrjiSoc; ixaaxov exaaxou iadxig 
TtoXXanXdaiov Xeyco, oxi oaaitXdaiov laxi xo AB xou E, 
xoaauxanXdaia scroti xal xd AB, TA xwv E, Z. 

EtceI ydp Eadxic ecm KoXXa^Xdoiov xo AB xou E xal 
xo TA xou Z, oaa dpa eaxlv ev xG AB ^.eyea}/] iaa x£> E, 
xoaauxa xal ev xw TA Taa xo Z. 8ir)pf|crd« xo [lev AB eic xd 
to E (ieye'dr) laa xd AH, HB, xo 8e TA eic. xd t£> Z laa xd 
TO, 0A- saxai 8r) laov xo TiXfj-dog xfiv AH, HB t£> TtXr|iL>£i 
tGv T6, 9A. xal STieUaov eotI to [ism AH t« E, to 8e T6 
tw Z, '(gov dpa to AH t« E, xal Ta AH, TO toTc E, Z. Sloc 
Ta auTa Sr] laov sotI to HB tw E, xal Ta HB, 0A tou; E, 
Z- oaa dpa eotIv ev to AB I'oa t£> E, ToaauTa xal ev toTc 
AB, TA I'aa toTc E, Z- ooaTiXdoiov dpa lav. to AB tou E, 
TooauTauXdoia saTai xal Ta AB, TA tcov E, Z. 

'Edv dpa f) oTiooaouv ^sysdr) otiooovouv [leyei&Gv 
I'owv to nkfjdoz exaoTov sxdoTou iadxic noXXanXdaiov, 
oaaitXdaiov eaTiv ev tcov jisys'daiv evoc, TooauTanXdoia 
eaTai xal Ta ndvTa twv itdvTiov oitsp e8ei SeT^ai. 



t In modern notation, this proposition reads m a + m f3 H 

P'- 

'Edv upcoTov SeuTepou iadxic f) uoXXanXdoiov xal Tprcov 
TeTapTou, f] 8e xal kc^ititov SeuTepou iadxic TtoXXaitXdaiov 
xal Ixtov TCTapTou, xal auvTSiDev TtpCrcov xal TtejiTtTov 
SeuTepou iadxic saTai noXXaTiXdoiov xal Tprcov xal sxtov 

TSTdpTOU. 

np«Tov ydp to AB BeuTEpou tou T iadxic soto tcoX- 
XanXdoiov xal Tphov to AE TSTapTou toO Z, egtw 8e xal 
ne[LKTOv to BH SeuTepou tou T iadxic TtoXXaTtXdaiov xal 
sxtov to EG TSTapTou tou Z - Xeyw, oti xal auvTEiDev 
npfiTov xal TCEjiTtTov to AH SeuTepou tou T iadxic eaxoa 
TioXXanXdaiov xal TpiTov xal extov to A9 TETapTou tou Z. 



times as one of the (first) magnitudes is (divisible) by 
one (of the second), so many times will all (of the first 
magnitudes) also (be divisible) by all (of the second) . 

A G B C H D 

i 1 1 i 1 1 

Ei 1 F' 1 

Let there be any number of magnitudes whatsoever, 
AB, CD, (which are) equal multiples, respectively, of 
some (other) magnitudes, E, F, of equal number (to 
them) . I say that as many times as AB is (divisible) by E, 
so many times will AB, CD also be (divisible) by E, F. 

For since AB, CD are equal multiples of E, F, thus 
as many magnitudes as (there) are in AB equal to E, so 
many (are there) also in CD equal to F. Let AB have 
been divided into magnitudes AG, GB, equal to E, and 
CD into (magnitudes) CH, HD, equal to F. So, the 
number of (divisions) AG, GB will be equal to the num- 
ber of (divisions) CH, HD. And since AG is equal to E, 
and CH to F, AG (is) thus equal to E, and AG, CH to E, 
F. So, for the same (reasons), GB is equal to E, and GB, 
HD to E, F. Thus, as many (magnitudes) as (there) are 
in AB equal to E, so many (are there) also in AB, CD 
equal to E, F. Thus, as many times as AB is (divisible) 
by E, so many times will AB, CD also be (divisible) by 
E, F. 

Thus, if there are any number of magnitudes what- 
soever (which are) equal multiples, respectively, of some 
(other) magnitudes, of equal number (to them), then as 
many times as one of the (first) magnitudes is (divisi- 
ble) by one (of the second), so many times will all (of the 
first magnitudes) also (be divisible) by all (of the second) . 
(Which is) the very thing it was required to show. 



Proposition X 

If a first (magnitude) and a third are equal multiples 
of a second and a fourth (respectively), and a fifth (mag- 
nitude) and a sixth (are) also equal multiples of the sec- 
ond and fourth (respectively), then the first (magnitude) 
and the fifth, being added together, and the third and the 
sixth, (being added together), will also be equal multiples 
of the second (magnitude) and the fourth (respectively) . 

For let a first (magnitude) AB and a third DE be 
equal multiples of a second C and a fourth F (respec- 
tively). And let a fifth (magnitude) BG and a sixth EH 
also be (other) equal multiples of the second C and the 
fourth F (respectively) . I say that the first (magnitude) 
and the fifth, being added together, (to give) AG, and the 
third (magnitude) and the sixth, (being added together, 



= m{a + 13 + ■■■). 



132 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



A B H 

i 1 1 1 1 1 

r i — i 

A E 

i 1 1 1 1 1 

Z' 1 

'Etc! yap iodxu; eax! TtoXXaitXdoiov to AB xou T xa! to 
AE toO Z, ooa dpa eoxlv ev xG AB Toa xG T, xooauxa xal 
ev xG AE 10a xG Z. Sia xd auxd Sr) xal ooa eoxlv ev iS BH 
I'aa xG T, xooauxa xal ev xG EG 10a xG Z- ooa dpa eoxlv ev 
6X9 xG AH 10a xG T, xooauxa xal ev oXtp xG A9 10a xG Z- 
ooarcXdoiov dpa eoxl xo AH xou T, xooauxarcXdoiov eoxai 
xal xo AQ xou Z. xal ouvxei9ev dpa TtpGxov xal tc^itixov xo 
AH Seuxepou xou T iodxu; eoxai TtoXXaitXdoiov xal xpixov 
xal exxov xo A6 xexdpxou xou Z. 

'Edv dpa TipGxov Seuxepou lodxic; fj TioXXarcXaoiov xal 
xpixov xexdpxou, fj Se xal nejiTixov Seuxepou iodxu; ttoX- 
XanXdoiov xal exxov xexdpxou, xal ouvxcdev TtpGxov xal 
tc^tixov Seuxepou lodxic; eoxai TioXXaTtXdoiov xal xpixov xal 
exxov xexdpxou- oitep eSei Sel^ai. 



t In modern notation, this propostion reads m a + n a = (m + n) a. 

Y- 

'Edv TipGxov Seuxepou lodxic; fj TioXXaiiXdoiov xal xpixov 
xexdpxou, X/]cpi9fj Se lodxic; TioXXaiiXdoia xou xe upGxou 
xal xpixou, xal Si' loou xGv Xrjqj'devxwv exdxepov exaxepou 
lodxic; eoxai TioXXajiXdaiov xo [Lev xou Seuxepou xo Se xou 
xexdpxou. 

npGxov Y a P TO A Seuxepou xou B lodxic; eoxw tioX- 
XajiXdaiov xal xpixov xo T xexdpxou xou A, xal eiXr]cp , dw 
xGv A, T lodxic; uoXXaTiXdaia xa EZ, H0- Xeyw, oxi lodxic; 
eoxl TioXXauXdoiov xo EZ xou B xal xo H9 xou A. 

'Etc! yap lodxic; eoxl KoXXanXdaiov xo EZ xou A xal 
xo H0 xou r, ooa dpa eoxlv ev xG EZ loa xG A, xooauxa 
xal ev xG H9 I'oa xG T. Sir)pf]o , dw xo ^ev EZ zlc, xd xG A 
^leyei!)/] I'oa xd EK, KZ, xo Se H0 zlc, xd xG T i'oa xa HA, 



to give) DH, will also be equal multiples of the second 
(magnitude) C and the fourth F (respectively) . 

A B G 

i 1 1 1 1 1 

C' — ' 

D EH 

i 1 1 1 1 1 

F i 1 

For since AB and DE are equal multiples of C and F 
(respectively), thus as many (magnitudes) as (there) are 
in AB equal to C, so many (are there) also in DE equal to 
F. And so, for the same (reasons), as many (magnitudes) 
as (there) are in BG equal to C, so many (are there) 
also in EH equal to F. Thus, as many (magnitudes) as 
(there) are in the whole of AG equal to C, so many (are 
there) also in the whole of DH equal to F. Thus, as many 
times as AG is (divisible) by C, so many times will DH 
also be divisible by F. Thus, the first (magnitude) and 
the fifth, being added together, (to give) AG, and the 
third (magnitude) and the sixth, (being added together, 
to give) DH, will also be equal multiples of the second 
(magnitude) C and the fourth F (respectively) . 

Thus, if a first (magnitude) and a third are equal mul- 
tiples of a second and a fourth (respectively), and a fifth 
(magnitude) and a sixth (are) also equal multiples of the 
second and fourth (respectively), then the first (magni- 
tude) and the fifth, being added together, and the third 
and sixth, (being added together), will also be equal mul- 
tiples of the second (magnitude) and the fourth (respec- 
tively) . (Which is) the very thing it was required to show. 



Proposition 3 f 

If a first (magnitude) and a third are equal multiples 
of a second and a fourth (respectively), and equal multi- 
ples are taken of the first and the third, then, via equality, 
the (magnitudes) taken will also be equal multiples of the 
second (magnitude) and the fourth, respectively. 

For let a first (magnitude) A and a third C be equal 
multiples of a second B and a fourth D (respectively), 
and let the equal multiples EF and GH have been taken 
of A and C (respectively). I say that EF and GH are 
equal multiples of B and D (respectively). 

For since EF and GH are equal multiples of A and 
C (respectively), thus as many (magnitudes) as (there) 
are in EF equal to A, so many (are there) also in GH 



133 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



AO- eaxai 8r] I'aov to Tikf(Qoc, tcov EK, KZ t£3 7iXr|iL>£i xfiv 
HA, AO. xal inel lodxic; eaxl TioXXaTrXdaiov to A tou B xal 
to T tou A, laov 8e to [ie\ EK t« A, to 8s HA tG T, 
iadxic; dpa taxi TroXXaTtXaaiov to EK tou B xal to HA tou 
A. Sid Ta auTa 8/) iadxi<; sotl TroXXaTrXdaiov to KZ tou B 
xal to A9 tou A. end ouv npfiTov to EK 8euTepou tou B 
I'odxic; eotI TroXXaTtXaaiov xal TpiTov to HA TSTapTou tou 
A, ecrci 8e xal nejiTCTov to KZ SeuTepou tou B lodxig TroX- 
XaTtXaaiov xal extov to A6 TSTapTou tou A, xal auvTE^ev 
dpa upcdTov xal TtefiTrcov to EZ SsuTspou tou B iadxic; eotI 
TroXXaTtXaaiov xal TpiTov xal sxtov to HO TETapTou tou A. 



A' 1 1 1 

B 1 

E K Z 

i 1 1 

ri — i — i — i 

A' — ' 

HA© 

i 1 1 

Edv dpa TtpCrtov BeuTEpou laaxic; fj TroXXaTtXaaiov 
xal TpiTov TSTapTou, X/]cpi9fj 8e tou TtpcVtou xal TpiTou 
iadxic; TtoXXaTtXdaia, xal 8i° I'aou tGv X/jcpi&svTiov sxaTspov 
sxaTspou iadxic; eaTai TroXXaTtXaaiov to tou 8euTepou 
to 8s tou TSTapTou- oTtep SSei 8eic;ai. 

t In modern notation, this proposition reads m(n a) = (m n) a. 

5'. 

'Edv 7ip£>T0v Ttpoc; 6euTepov tov auTov zyr\ Xoyov xal 
Tpaov Ttpoc; TETapTov, xal Ta iadxu; TtoXXaitXdaia tou te 
upcoTou xal TpiTou Ttpoc; Ta iadxic; TtoXXaTtXdaia tou 8euTepou 
xal TSTapTou xai}° ottoiovouv TtoXXaTtXaaiaajjiov tov auTov 
e^ei Xoyov Xrjcp^evTa xaTaXXrjXa. 

npfiTov yap to A Ttpoc; SeuTepov to B tov auTov eyixto 
Xoyov xal xpixov to T Ttpoc; xexapxov to A, xal eiXrjtp'dw 
xfiv [ie\ A, T iadxic; TtoXXaTtXdaia Ta E, Z, t«v 8e B, A 
dXXa, a exuxev, iadxic; TtoXXaTtXdaia Ta H, 6- Xsyw, oxi 
saxlv cbc; to E Ttpoc; to H, ouxcoc; to Z Ttpoc; to 6. 



equal to C. Let EF have been divided into magnitudes 
EK, KF equal to A, and GH into (magnitudes) GL, LH 
equal to C. So, the number of (magnitudes) EK, KF 
will be equal to the number of (magnitudes) GL, LH. 
And since A and C are equal multiples of B and D (re- 
spectively), and EK (is) equal to A, and GL to C, EK 
and GL are thus equal multiples of B and D (respec- 
tively). So, for the same (reasons), KF and LH are equal 
multiples of B and D (respectively) . Therefore, since the 
first (magnitude) EK and the third GL are equal mul- 
tiples of the second B and the fourth D (respectively), 
and the fifth (magnitude) KF and the sixth LH are also 
equal multiples of the second B and the fourth D (re- 
spectively), then the first (magnitude) and fifth, being 
added together, (to give) EF, and the third (magnitude) 
and sixth, (being added together, to give) GH, are thus 
also equal multiples of the second (magnitude) B and the 
fourth!) (respectively) [Prop. 5.2]. 

A I 1 1 1 

B' 1 

E K F 

i 1 1 

C i — " — i — i 
D' 1 

G L H 

i 1 1 

Thus, if a first (magnitude) and a third are equal mul- 
tiples of a second and a fourth (respectively), and equal 
multiples are taken of the first and the third, then, via 
equality, the (magnitudes) taken will also be equal mul- 
tiples of the second (magnitude) and the fourth, respec- 
tively. (Which is) the very thing it was required to show. 



Proposition 4 f 

If a first (magnitude) has the same ratio to a second 
that a third (has) to a fourth then equal multiples of the 
first (magnitude) and the third will also have the same 
ratio to equal multiples of the second and the fourth, be- 
ing taken in corresponding order, according to any kind 
of multiplication whatsoever. 

For let a first (magnitude) A have the same ratio to 
a second B that a third C (has) to a fourth D. And let 
equal multiples E and F have been taken of A and C 
(respectively), and other random equal multiples G and 



134 



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ELEMENTS BOOK 5 



A' 1 

B i-H 

Ei 1 1 

H 1 — i — i — 1 

K' 1 ' 

Mi 1 1 1 

r i — i 
Zi — i — i 

A, , , 

N' ' ' ' 

EiXrjcp'dco yap xfiiv (lev E, Z iadxu; TtoXXanXdaia xd K, 
A, t«v 5e H, dXXa, a exuxev, iadxig TioXXanXdoia xd M, 
N. 

[Kod] fejtel tadxig eaxl TtoXXanXdaiov to jisrv E xou A, xo 
8e Z xou r, xal eiXr)7ixai xfiv E, Z I'odxic; noXXanXdoia xd K, 
A, ladxic dpa sraxl noXXanXdoiov xo K xou A xal xo A xou 
T. Bid xd auxd 8r| ladxi? eaxl KoXXanXdaiov xo M xou B xal 
xo N xou A. xal inei eaxiv cbc xo A upoc; xo B, oux«<; xo T 
7ip6<; xo A, xal EiXrjTtxai xov ^tev A, T ladxic; TtoXXaTtXdaia 
xd K, A, xQv 8e B, A dXXa, a exu)(£v, iadxu; TtoXXaTtXdaia 
xd M, N, ei dpa UTtepe)(£i xo K xou M, UTtepexei xocl xo A 
xou N, xal si laov, laov, xal et eXaxxov, eXaxxov. xal saxi 
xd [Ltv K, A xQv E, Z iadxi<; noXXauXdaia, xd 8e M, N xwv 
H, dXXa, a exu)(£v, iadxic; noXXaKXaaia' saxiv dpa foe, xo 
E Ttpoc xo H, oux«<; xo Z Tipoc xo 0. 

'Edv dpa upwxov Kpbc, Ssuxepov xov auxov i)(T] Xoyov 
xal xpixov 7tp6<; xexapxov, xal xd ladxu; noXXanXdaia xou xe 
Tipoxou xal xptxou npbz xd ladxu; TioXXauXdaia xou Seuxepou 
xal xexdpxou xov auxov e^ei Xoyov xai}' okoiovouv uoXXa- 
TiXaaiao^tov XTjcpiJevxa xaxdXXrjXa- ojisp eSsi 8el5ai. 



H of B and D (respectively) . I say that as E (is) to G, so 
F (is) to H. 

A ' ' 

B i—i 

E i 1 1 

G i — i — i — i 

K i 1 1 

Mi 1 1 1 

C i — i 
D ' ' 

F i 1 1 

H 

L i 1 1 

N i 1 1 1 

For let equal multiples K and L have been taken of E 
and F (respectively), and other random equal multiples 
M and N of G and H (respectively) . 

[And] since E and F are equal multiples of A and 
C (respectively), and the equal multiples K and L have 
been taken of E and F (respectively), K and L are thus 
equal multiples of A and C (respectively) [Prop. 5.3]. So, 
for the same (reasons), M and N are equal multiples of 
B and D (respectively). And since as A is to B, so C (is) 
to D, and the equal multiples K and L have been taken 
of A and C (respectively), and the other random equal 
multiples M and N of B and D (respectively), then if K 
exceeds M then L also exceeds N, and if (K is) equal (to 
M then L is also) equal (to N), and if {K is) less (than M 
then L is also) less (than N) [Def. 5.5]. And K and L are 
equal multiples of E and F (respectively), and M and N 
other random equal multiples of G and H (respectively) . 
Thus, as E (is) to G, so F (is) to H [Def. 5.5]. 

Thus, if a first (magnitude) has the same ratio to a 
second that a third (has) to a fourth then equal multi- 
ples of the first (magnitude) and the third will also have 
the same ratio to equal multiples of the second and the 
fourth, being taken in corresponding order, according to 
any kind of multiplication whatsoever. (Which is) the 
very thing it was required to show. 



t In modern notation, this proposition reads that if a : (3 : : 7 : $ then m a : n /3 : : m 7 : n <5, for all m and n. 



135 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



s . 

'Eav [liyE'doz (leye'doug iadxic; fj TtoXXaTtXdaiov, onep 
dcpaips-dev dcpaips-devToc;, xal to Xolttov tou Xomou iadxic; 
soxai TroXXaTtXdcHov, oacmXdmov soti to oXov tou oXou. 

A E B 

i 1 1 1 — i — i — i 

H r Z A 

I 1 1 1 

Meye'doc yap to AB [leye-Qouc, tou IA iadxi<; sgtco TtoX- 
XoaiXdaiov, orcep d(paipe , d£v to AE dcpaipe'dEVToc; tou TZ' 
Xeyw, oti xal Xoitiov to EB Xoitiou tou ZA Eadxic; eaxoa 
TtoXXaitXaaiov, oaomXdaiov eaTiv oXov to AB oXou tou IA. 

'OaanXdaiov ydp taxi to AE tou TZ, ToaauTaaiXdaiov 
ysyovsTW xal to EB tou TH. 

Kal snel ladxn; ecttI TtoXXaTtXdaiov to AE tou TZ xal to 
EB tou Hr, iadxi<; apa sotI TtoXXaTtXdaiov to AE tou TZ 
xal to AB tou HZ. xeroxi Be iadxii; uoXXaTiXdoiov to AE 
tou TZ xal to AB tou IA. ladxig apa sgti noXXauXdaiov to 
AB ExaTepou twv HZ, EA igov apa to HZ iu IA. xoivov 
a(pr}p7)a-Q(x> to TZ- Xoiitov apa to Hr Xomw tc5 ZA igov 
eaTiv. xal em\ iadxi<; laxl TtoXXaTtXdaiov to AE tou TZ 
xal to EB tou Hr, I'aov 8e to Hr iu AZ, ladxic; apa sgtI 
TioXXauXdaiov to AE tou TZ xal to EB tou ZA. ladxic 8s 
UTioxeiTai TtoXXaTtXdaiov to AE tou TZ xal to AB tou EA 
iadxi<; apa eotI TtoXXaTtXdaiov to EB tou ZA xal to AB 
tou TA. xal XoiTtov apa to EB Xoitiou tou ZA iadxic; eaTai 
TtoXXaTtXdaiov, oaaTtXdaiov eaTiv oXov to AB oXou tou TA. 

'Eav apa [leye'doz [leye'douci iadxn; fj TtoXXaTtXdaiov, 
OTtep dcpaipcdev dcpaipcdevToc;, xal to XoiTtov tou XoiTtoO 
iadxic eaTai TtoXXaTtXdaiov, oaaTtXdaiov eaTi xal to oXov 
tou oXou - OTtep e5ei 8elc;ai. 



t In modern notation, this proposition reads ma - m fS = m (a - fS). 

f'. 

'Eav 8uo jieyei&r) 5uo [ieycOtov iadxic fj TtoXXaTtXdaia, 
xal dcpaipeiJevTa Tivd twv auTWv iadxic fj TtoXXaTtXdaia, xal 
Ta XoiTia toTc auTolc f]Toi laa eaTiv fj iadxic auT«v TtoX- 
XaTtXdaia. 

Auo yap jieyei&r) Ta AB, TA 8uo ^teycdCSv xfiv E, Z 



Proposition 5* 

If a magnitude is the same multiple of a magnitude 
that a (part) taken away (is) of a (part) taken away (re- 
spectively) then the remainder will also be the same mul- 
tiple of the remainder as that which the whole (is) of the 
whole (respectively). 

A E B 

i 1 1 1 1 1 — i 

G C F D 

i — i 1 1 

For let the magnitude AB be the same multiple of the 
magnitude CD that the (part) taken away AE (is) of the 
(part) taken away CF (respectively). I say that the re- 
mainder EB will also be the same multiple of the remain- 
der FD as that which the whole AB (is) of the whole CD 
(respectively). 

For as many times as AE is (divisible) by CF, so many 
times let EB also have been made (divisible) by CG. 

And since AE and EB are equal multiples of CF and 
GC (respectively), AE and AB are thus equal multiples 
of CF and GF (respectively) [Prop. 5.1]. And AE and 
AB are assumed (to be) equal multiples of CF and CD 
(respectively). Thus, AB is an equal multiple of each 
of GF and CD. Thus, GF (is) equal to CD. Let CF 
have been subtracted from both. Thus, the remainder 
GC is equal to the remainder FD. And since AE and 
EB are equal multiples of CF and GC (respectively), 
and GC (is) equal to DF, AE and EB are thus equal 
multiples of CF and FD (respectively). And AE and 
AB are assumed (to be) equal multiples of CF and CD 
(respectively). Thus, EB and AB are equal multiples of 
FD and CD (respectively). Thus, the remainder EB will 
also be the same multiple of the remainder FD as that 
which the whole AB (is) of the whole CD (respectively) . 

Thus, if a magnitude is the same multiple of a magni- 
tude that a (part) taken away (is) of a (part) taken away 
(respectively) then the remainder will also be the same 
multiple of the remainder as that which the whole (is) of 
the whole (respectively) . (Which is) the very thing it was 
required to show. 



Proposition 6 f 

If two magnitudes are equal multiples of two (other) 
magnitudes, and some (parts) taken away (from the for- 
mer magnitudes) are equal multiples of the latter (mag- 
nitudes, respectively), then the remainders are also either 
equal to the latter (magnitudes), or (are) equal multiples 



136 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



ladxu; eaxw TtoXXaTtXdaia, xal dcpaipedevxa xd AH, TO xuv 
auxwv xGv E, Z ladxu; eaxw TtoXXaTtXdaia- Xeyw, oxi xdi 
Xomd xd HB, 0A xoTg E, Z f]xoi laa eaxiv f] Eadxic; auxcov 
TtoXXaTtXdaia. 



A H B 

i 1 1 1 1 

Ei 1 

k r © a 

I 1 1 1 1 1 

Z' — ' 

TEaxco yap Ttpoxepov xo HB iw E l'aov Xeyw, oxi xdi 
xo 0A t& Z i'aov eaxiv. 

Keicrda) ydp xw Z iaov xo TK. eTtel iadxi<; eaxl TtoX- 
XaTtXdaiov xo AH xou E xdi xo T9 xou Z, i'aov 8e xo ^iev HB 
iG E, xo 8e KT xo Z, ladxi? dpa eaxl TtoXXaTtXdaiov xo AB 
xou E xdi xo K6 xou Z. ladxu; 8e UTtoxeixai TtoXXaTtXdaiov 
xo AB xou E xdi xo LA xou Z - ladxi? dpa eaxl TtoXXaTtXdaiov 
xo K9 xou Z xdi xo TA xou Z. etc! ouv exdxepov xGv K9, 
TA xou Z tadxu; eaxl TtoXXaTtXdaiov, i'aov dpa eaxl xo K0 
iS TA. xoivov dcpr)pr]a , do xo T0 - Xoitiov dpa xo Kr Xomw 
xo OA laov eaxiv. dXXd xo Z iw Kr eaxiv taov xal xo 
6A dpa xw Z Taov eaxiv. waxe ei xo HB xo E laov eaxiv, 
xal xo OA laov eaxai xw Z. 

'Ojioioi; 8r) 5e(?o^.ev, oxi, xav uoXXaTiXdaiov f) xo HB 
xou E, xoaauxaTiXdaiov eaxai xal xo 6A xou Z. 

Edv dpa 8uo ^.eyeiDr] 8uo ^eye'dQv ladxu; f) 710X- 
XaiiXdaia, xal dcpaipeiDevxa xivd xwv auxov ladxic f) 710X- 
XaiiXdaia, xal xa Xomd xoT<; auxolc f]xoi laa eaxiv f] iadxn; 
auxCSv TioXXauXdaia- oTiep e8ei SeT^ai. 



t In modern notation, this proposition reads ma - na = (m - n) a. 

c 

Td iaa npbc, xo auxo xov auxov exei Xoyov xal xo auxo 
izpoq xd laa. 

TSaxw I'aa ^teye'dr] xd A, B, aXXo 8e xi, o exuxev, 
^eyeiDot; xo T- Xeyw, oxi exdxepov xwv A, B 7ip6<; xo T 
xov auxov e)(ei Xoyov, xal xo T up6<; exdxepov xwv A, B. 



of them (respectively) . 

For let two magnitudes AB and CD be equal multi- 
ples of two magnitudes E and F (respectively) . And let 
the (parts) taken away (from the former) AG and CH be 
equal multiples of E and F (respectively) . I say that the 
remainders GB and HD are also either equal to E and F 
(respectively), or (are) equal multiples of them. 

A G B 

1 1 1 1 1 

Ei ' 

K C H D 

1 1 1 1 1 1 

Fi ' 

For let GB be, first of all, equal to E. I say that HD is 
also equal to F. 

For let CK be made equal to F. Since AG and CH 
are equal multiples of E and F (respectively), and GB 
(is) equal to E, and KG to F, AB and KH are thus equal 
multiples of E and F (respectively) [Prop. 5.2]. And AB 
and CD are assumed (to be) equal multiples of E and F 
(respectively). Thus, KH and CD are equal multiples of 
F and F (respectively). Therefore, KH and CD are each 
equal multiples of F. Thus, KH is equal to CD. Let CH 
have be taken away from both. Thus, the remainder KG 
is equal to the remainder HD. But, F is equal to KG. 
Thus, HD is also equal to F. Hence, if GB is equal to E 
then HD will also be equal to F. 

So, similarly, we can show that even if GB is a multi- 
ple of E then HD will also be the same multiple of F. 

Thus, if two magnitudes are equal multiples of two 
(other) magnitudes, and some (parts) taken away (from 
the former magnitudes) are equal multiples of the latter 
(magnitudes, respectively), then the remainders are also 
either equal to the latter (magnitudes), or (are) equal 
multiples of them (respectively). (Which is) the very 
thing it was required to show. 



Proposition 7 

Equal (magnitudes) have the same ratio to the same 
(magnitude), and the latter (magnitude has the same ra- 
tio) to the equal (magnitudes) . 

Let A and B be equal magnitudes, and C some other 
random magnitude. I say that A and B each have the 



137 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



A 1 1 A 1 1 1 1 1 

B 1 1 E 1 1 1 1 1 

r> ' Z' ' ' ' 

EiXf|(p , v)w yap twv jiev A, B iadxu; TtoXXaTiXdaia xd A, 
E, toD 8e r dXXo, 6 stu)(sv, TtoXXauXdaiov to Z. 

EtceI ouv ladxic eaxl TtoXXaTtXdaiov to A tou A xal to 
E tou B, I'aov 8e to A tw B, laov dpa xal to A tw E. dXXo 
M, 6 ctu/cv, to Z. Ei dpa UTtepexei to A tou Z, UTtepexei 
xal to E tou Z, xal el Taov, laov, xal si eXaTTov, eXaTTov. 
xai eaTi Td [lev A, E twv A, B ladxic; TtoXXaitXdaia, to Be: 
Z tou r dXXo, o eTU^ev, TtoXXaTiXdaiov eotiv dpa w<; to A 
7ip6<; to T, outco? to B 7ipo<; to T. 

Asyw [8/|], oti xal to T 7tp6<; exaTepov twv A, B tov 
auTov e)(Ei Xoyov. 

Twv yap auTWv xaTaaxeuaa'devTWv 6|ioiw<; Bel^o^iev, 
oti laov ecru to A iw E- dXXo 8e ti to Z- ei dpa UTtepexei 
to Z tou A, UTiepexei xal tou E, xal el laov, I'aov, xal ei 
eXaTTov, eXaTTov. xai soti to [lev Z tou T TtoXXaTtXdaiov, 
Ta 8e A, E twv A, B aXXa, a stu^sv, ladxic; TtoXXauXdaia' 
eaTiv dpa w<; to T Ttp6<; to A, outwc to T npbc, to B. 

Ta I'aa dpa Kpo<; to auTo tov ai)Tov e^ei Xoyov xal to 
ai)TO Ttpoc; Ta loot. 



same ratio to C, and (that) C (has the same ratio) to 
each of A and B. 

A i 1 D 1 1 1 1 1 

B 1 1 E 1 1 1 1 1 

C 1 1 F i 1 1 1 

For let the equal multiples D and E have been taken 
of A and B (respectively), and the other random multiple 
FofC. 

Therefore, since D and E are equal multiples of A 
and B (respectively), and A (is) equal to B, D (is) thus 
also equal to E. And F (is) different, at random. Thus, if 
D exceeds F then E also exceeds F, and if (D is) equal 
(to F then E is also) equal (to F), and if (D is) less 
(than F then E is also) less (than F) . And D and E are 
equal multiples of A and f? (respectively), and F another 
random multiple of C. Thus, as A (is) to C, so B (is) to 
C [Def. 5.5]. 

[So] I say that C 1 * also has the same ratio to each of A 
and B. 

For, similarly, we can show, by the same construction, 
that D is equal to E. And F (has) some other (value). 
Thus, if F exceeds D then it also exceeds E, and if (F is) 
equal (to D then it is also) equal (to E), and if (F is) less 
(than D then it is also) less (than E) . And F is a multiple 
of C, and _D and E other random equal multiples of A 
and B. Thus, as C (is) to A, so C (is) to B [Def. 5.5]. 

Thus, equal (magnitudes) have the same ratio to the 
same (magnitude), and the latter (magnitude has the 
same ratio) to the equal (magnitudes). 



I16piG[Jia. 

Ex 8r) toutou cpavepov, oti edv (jLEye'Or] Tiva dvdXoyov 
fl, xal dvdrcaXiv dvdXoyov eaTai. oiiep s8ei 8eT<;ai. 

t The Greek text has "E", which is obviously a mistake. 

t In modern notation, this corollary reads that if a : fi : : 7 : 5 then (3 : a 



Corollary* 

So (it is) clear, from this, that if some magnitudes are 
proportional then they will also be proportional inversely. 
(Which is) the very thing it was required to show. 



:: 5 : 7. 



Twv dvlawv jieycdcov to \±ei^ov Ttp6<; to ai)To ^.el^ova 
Xoyov eyei f\nep to eXaTTov. xal to auTo npbc, to eXaTTov 
Xoyov sysi. fjjiep 7ipo<; to \leXC,o\. 

'EaTW dviaa ^teye'dr] Ta AB, T, xal sotw ^.el^ov to AB, 
dXXo 8e, o etu)(Ev, to A' Xeyw, otl to AB npbz to A 
^.el^ova Xoyov e/et ^itep to T npbq to A, xal to A npbz 
to T ^el^ova Xoyov ex ei W S P npbc, to AB. 



Proposition 8 

For unequal magnitudes, the greater (magnitude) has 
a greater ratio than the lesser to the same (magnitude) . 
And the latter (magnitude) has a greater ratio to the 
lesser (magnitude) than to the greater. 

Let AB and C be unequal magnitudes, and let AB be 
the greater (of the two), and D another random magni- 
tude. I say that AB has a greater ratio to D than C (has) 
to D, and (that) D has a greater ratio to C than (it has) 
to AB. 



138 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



A E 



B 



A 



E B 



A E 



B 



A 



E B 



r h 



K h 
A h 



H 



e 



H 



(-) 



H 



H 



K 
A 
A 



N ' 1 1 1 ' N 

Tkei yap [iei£6v eaxi to AB tou T, xeiai9a> to T laov 
to BE - to 8rj eXaaaov iSv AE, EB TroXXaTtXaaiaCo^iEvov 
ecrcoa ttots toO A ^eTi^ov. screw npoxepov to AE sXaTTov 
toO EB, xal KETioXXaTiXaaida'do to AE, xal eaTW auToO 
TioXXauXdaiov to ZH \±s%ov ov tou A, xal oaaitXdaiov saTi 
to ZH toO AE, ToaauTanXdaiov yEyovsTO xal to \±ev H9 
toO EB to Se K tou T- xal dXVjcp'da) toO A SmXdaiov jiev 
to A, TpiTiXdaiov 5e to M, xal Ec;fjc; svl TtXaov, av to 
Xa^pavo^tevov TioXXauXdaiov [lev yev/]Tai toO A, uptOTC*; 8e 
^b£ov tou K. eiXricp'dw, xal eoto to N TSTpauXdaiov jiev 
tou A, upwTwc; 8e ^ta^ov tou K. 

'Etc! ouv to K tou N TtpwTCOc; sotIv eXaTTov, to K dpa 
tou M oux eaTiv eXaTTov. xal euel iadxic; sotI TtoXXairXdaiov 
to ZH tou AE xal to H0 tou EB, iadxic; dpa taxi 710X- 
XaTtXdaiov to ZH tou AE xal to Z8 tou AB. iadxic; 8£ 
eaTi TtoXXanXdaiov to ZH tou AE xal to K tou T- iadxic; 
dpa taxi TioXXanXdaiov to ZO tou AB xal to K tou T. xa 
Z0, K dpa twv AB, T iadxic; eoxl TtoXXaTtXdaia. udXiv, end 
iadxic; eotI TtoXXauXdaiov to HO tou EB xal to K tou T, 
taov 8e to EB x& T, Taov dpa xal to H0 tw K. to Se K 
tou M oux eaTiv eXaTTov ou8' dpa to H0 tou M eXaTTov 
eaTiv. ^.eI^ov Be to ZH tou A' oXov dpa to Z0 auvaji- 
cpoTepwv twv A, M [Lel^ov eaTiv. dXXa auva^tcpoTcpa Ta A, 
M tw N eaTiv I'aa, £7iei8r]Ti£p to M tou A TpiTtXdaiov eaTiv, 
auva^icpoTspa 8s Ta M, A tou A taxi TCTpauXdaia, ecrci 8e 
xal to N tou A TSTpajiXdaiov auvajicpoTcpa dpa Ta M, A 
to N laa eaTiv. dXXa to ZO tgSv M, A \ieiZ,6v eaTiv to 
Z0 dpa tou N bizepiyev to 8e K tou N oux UTtepe/ei. xal 
taxi Ta (lev Z0, K t£>v AB, T iadxic; TioXXaTtXdaia, to 8e N 
tou A dXXo, o £TU)(ev, TioXXaTtXdaiov to AB dpa Tipoc; to 
A jiei^ova Xoyov e)(ei f\Ksp to T Tipoc; to A. 

Aeyco 8r|, oti xal to A Tipoc; to T ^lei^ova Xoyov s^si 
f\mp to A Tipoc; to AB. 

Twv yap auTQv xaTaaxeuaa'devTWv ojioiwc; Seic^o^iev, 
oti to ^tev N tou K (mepexei, to Se N tou Z9 oux UTiepexei. 
xai eaxi to fiev N tou A TioXXaTiXdaiov, Ta 8e Z6, K twv 
AB, r dXXa, a ctuxev, iadxic; TioXXaTiXdaia- to A dpa Tipoc; 
to T ^iei£ova Xoyov exei f]7iep to A Tipoc; to AB. 

AXXa Br] to AE tou EB ^la^ov eoto. to 8r) eXaTTov 
to EB TtoXXaTiXaaia£6|i£vov eaTai tcots tou A [ie%ov. ne- 



K ^ 
D > 
L ^ 



K ^ 
D > 
L ^ 



N i 1 1 1 1 N 1 1 1 1 1 

For since AB is greater than C, let be made equal 
to C. So, the lesser of AE and EB, being multiplied, will 
sometimes be greater than D [Def. 5.4]. First of all, let 
AE be less than EB, and let AE have been multiplied, 
and let FG be a multiple of it which (is) greater than 
D. And as many times as FG is (divisible) by AE, so 
many times let GH also have become (divisible) by EB, 
and K by C. And let the double multiple L of D have 
been taken, and the triple multiple M, and several more, 
(each increasing) in order by one, until the (multiple) 
taken becomes the first multiple of D (which is) greater 
than K. Let it have been taken, and let it also be the 
quadruple multiple N of D — the first (multiple) greater 
than K. 

Therefore, since K is less than A^ first, K is thus not 
less than M. And since FG and GH are equal multi- 
ples of AE and EB (respectively), FG and FH are thus 
equal multiples of AE and AB (respectively) [Prop. 5.1]. 
And FG and K are equal multiples of AE and C (re- 
spectively) . Thus, FH and K are equal multiples of AB 
and C (respectively). Thus, FH, K are equal multiples 
of AB, C. Again, since GH and K are equal multiples 
of EB and C, and EB (is) equal to C, GH (is) thus also 
equal to K. And K is not less than M. Thus, GH not less 
than M either. And FG (is) greater than D. Thus, the 
whole of FH is greater than D and M (added) together. 
But, D and M (added) together is equal to N, inasmuch 
as M is three times D, and M and D (added) together is 
four times D, and N is also four times D. Thus, M and D 
(added) together is equal to N. But, FH is greater than 
M and D. Thus, FH exceeds N. And K does not exceed 
N. And FH, K are equal multiples of AB, C, and N 
another random multiple of D. Thus, AB has a greater 
ratio to I? than C (has) to I? [Def. 5.7]. 

So, I say that D also has a greater ratio to C than D 
(has) to AB. 

For, similarly, by the same construction, we can show 
that N exceeds K, and N does not exceed FH. And 
A^ is a multiple of D, and FH, K other random equal 
multiples of AB, C (respectively) . Thus, D has a greater 



139 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



TtoXXarcXaaida'dM, xal eaxw to HO TtoXXaTtXdaiov \xz\ tou 
EB, jiei^ov 8e tou A- xal oaaitXdaiov tou to HO tou EB, 
ToaauTarcXdaiov yeyoveTCO xal to [lev ZH tou AE, to Se K 
tou r. b\ioicdci 8r) Bei^o^iev, oxi Ta ZO, K twv AB, T ladxic; 
eaTi TioXXaTtXdaia- xal eiXricp'dw 6[io(w<; to N TtoXXanXdaiov 
^terv tou A, TipwT«<; 5e fielCov tou ZH- uaie TidXiv to ZH 
tou M oux eaTiv sXaaaov. ^ieI^ov 8e to HO tou A - 6Xov 
apa to ZO twv A, M, toutegti. tou N, hizepe-^ei. to 8s K 
tou N oux uneps/si, en:£i8r]Tiep xal to ZH ^el^ov ov tou 
HO, toutsoti tou K, tou N oux UTtEpe/ei. xal (i>aauT(x><; 
xaTaxoXou-douvTsg tou; £Tidv« Tiepaivojisv ttjv dnoSsi^iv. 

Tcov apa dviaiov ^teyer'dwv to jisI^ov Tipoc; to auTO 
(lei^ova Xoyov E)(£i fjjiEp to eXaTTov xal to ain6 Ttpoc; to 
eXaTTOv [idi^ova Xoyov s)(ei rjnep iipo? to ^ei£ov oitep sSei 
8eT^ai. 



ft'. 

Ta Tipog to auTO tov auTov e)(ovTa Xoyov laa dXXrjXou; 
sotiv xal 7ipo<; a to aura tov auTov e)(ei Xoyov, exelva laa 

SGTIV. 

A' 1 Bi 1 

r> 1 

'E)(£tw yap exaTEpov xwv A, B rcpoc; to T tov auTov 
Xoyov Xeyco, oti Taov ecru, to A tw B. 

EE yap \ir\, oux av exaTepov twv A, B 7tp6c to T tov 
auTov elx £ Xoyov £)(£i 8e- 1'aov apa eotI to A iu B. 

'E)(£tcl) 8r) udXiv to r 7ip6<; exaTepov twv A, B tov auTov 
Xoyov Xsyco, oti laov eaTi to A tw B. 

El yap oux av to T npoc, exaTspov xSv A, B tov 
auTov el)(£ Xoyov £)(ei 8e- i'aov apa eotI to A iu B. 

Ta apa 7tp6<; to auTO tov auTov £)(ovTa Xoyov laa 
dXXfjXou; law xal 7tp6<; a to auTO tov auTov £)( £l Xoyov, 
exelva laa eaTiv oitep eBei SeTc;ai. 



l . 

Tcov npoc, to auTO Xoyov £)(6vt«v to jiei^ova Xoyov 
e/ov exelvo ^tsT^ov caw 7tp6<; o 8e to ai)To ^ici^ova Xoyov 



ratio to C than D (has) to ^4_B [Def. 5.5]. 

And so let AE be greater than EB. So, the lesser, 
EB, being multiplied, will sometimes be greater than D. 
Let it have been multiplied, and let GH be a multiple of 
EB (which is) greater than D. And as many times as 
GH is (divisible) by EB, so many times let FG also have 
become (divisible) by AE, and K by C. So, similarly 
(to the above), we can show that FH and K are equal 
multiples of AB and C (respectively) . And, similarly (to 
the above), let the multiple of D, (which is) the first 
(multiple) greater than FG, have been taken. So, FG 
is again not less than M. And GH (is) greater than D. 
Thus, the whole of FH exceeds D and M, that is to say 
N. And K does not exceed N, inasmuch as FG, which 
(is) greater than GH — that is to say, K — also does not 
exceed N. And, following the above (arguments), we 
(can) complete the proof in the same manner. 

Thus, for unequal magnitudes, the greater (magni- 
tude) has a greater ratio than the lesser to the same (mag- 
nitude) . And the latter (magnitude) has a greater ratio to 
the lesser (magnitude) than to the greater. (Which is) the 
very thing it was required to show. 

Proposition 9 

(Magnitudes) having the same ratio to the same 
(magnitude) are equal to one another. And those (mag- 
nitudes) to which the same (magnitude) has the same 
ratio are equal. 

A i 1 B 1 

C' 1 

For let A and B each have the same ratio to C. I say 
that A is equal to B. 

For if not, A and B would not each have the same 
ratio to C [Prop. 5.8]. But they do. Thus, A is equal to 
B. 

So, again, let C have the same ratio to each of A and 
B. I say that A is equal to B. 

For if not, C would not have the same ratio to each of 
A and B [Prop. 5.8]. But it does. Thus, A is equal to B. 

Thus, (magnitudes) having the same ratio to the same 
(magnitude) are equal to one another. And those (magni- 
tudes) to which the same (magnitude) has the same ratio 
are equal. (Which is) the very thing it was required to 
show. 

Proposition 10 

For (magnitudes) having a ratio to the same (mag- 
nitude), that (magnitude which) has the greater ratio is 



140 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



£/£i, exelvo eXaTTov eaTiv. 

A' 1 Bi 1 

r< 1 

'E^etco yap to A Trpoc; to V jiei^ova Xoyov fjicep to B 
iipot; to r- Xeyto, oti [icTCov eaTi to A tou B. 

Et yap \jx\, fjxoi Taov eaTi to A i« B ij eXaaaov. iaov 
\iev ouv oux cotI to A tw B' exaTepov yap av twv A, B 
7ip6<; to r tov auTov eT/s Xoyov. oux exei 8e' oux apa iaov 
eaTi to A tu B. ou8e [ir]-v eXaaaov ecra to A tou B- to A 
yap av Tipoc, to T eXdaaova Xoyov sl/ev f]Trsp to B Trpoc. to 
T. oux e^ei Se' oux apa eXaaaov 6cm to A tou B. eSsi^i!)/] 
8s ou5e loov jieTCov apa eaTi to A tou B. 

'E^stw 8rj TidXiv to T npbc, to B jiei^ova Xoyov f]7rep to 
r Ttpoc. to A' Xeyw, oti eXaaaov coti to B tou A. 

El yap y.r\, f]Toi iaov eaTiv f\ y.e%ov. Iaov ^tev ouv oux 
eaTi to B t« A- to T yap dv rcpoc. exaTepov t«v A, B tov 
auTov eT)(£ Xoyov. oux e^ei 56' °'- ,x "P a ' aov £ax ' 1 T ° A 
tu B. ouSe u/]v ^el£6v eaTi to B tou A- to T yap av TTpoc, 
to B eXdaaova Xoyov eTxev r\mp npbz to A. oux ex £l 5e' 
oux apa pie'iCov eaTi to B tou A. eBeix^r) Se, oti ou5e iaov 
eXaTTov apa eaTi to B tou A. 

Twv apa irpog to ai)To Xoyov cxovtwv to [ici^ova Xoyov 
exov (jieTCov eaTiv xal Trpoc, o to auTO jiei^ova Xoyov exei, 
exeTvo eXaTTov eaTiv oirep eSei SeT^ai. 



ia'. 

Oi tw auTW Xoyco ol auTol xal dXXiqXoic, eialv oi auToi. 

A i — i r ' ' E' ' 

B' — i A 1 Z 1 

H 1 1 1 ® 1 1 1 K 1 1 1 

Ai — i — i — i M 1 1 1 N 1 1 1 1 

'EaTwaav yap (be; ^iev to A 7rpo<; to B, outo<; to T npbc, 

to A, (b<; 6e to r Ttpog to A, outwi; to E 7ipo<; to Z - Xeya>, 

oti eaTiv (bg to A 7ipo<; to B, outoc to E irpoc; to Z. 

EiXrjcp'dco yap tuv A, T, E Eadxi? iroXXaTiXdaia Ta H, 0, 

K, t(5v 5e B, A, Z aXXa, a stuxsv, Eadxi? TroXXaTiXdaia Ta 

A, M, N. 

Kal ettel eaTiv (be, to A irpoc to B, outgjc; to T npoc to 
A, xal eiXrjTTTai tSv [lev A, T ladxig noXXaTiXdaia Ta H, 0, 
twv 8e B, A aXXa, a eTUxev, ladxic, TioXXairXdaia Ta A, M, 
ei apa uirepexei to H tou A, UTiepexei xal to tou M, xal ei 
iaov eaTiv, Iaov, xal ei eXXemei, eXXemei. TidXiv, cttci eaTiv 



(the) greater. And that (magnitude) to which the latter 
(magnitude) has a greater ratio is (the) lesser. 

A i 1 B 1 

C' 1 

For let A have a greater ratio to C than B (has) to C. 
I say that A is greater than B. 

For if not, A is surely either equal to or less than B. 
In fact, A is not equal to B. For (then) A and B would 
each have the same ratio to C [Prop. 5.7]. But they do 
not. Thus, A is not equal to B. Neither, indeed, is A less 
than B. For (then) A would have a lesser ratio to C than 
B (has) to C [Prop. 5.8]. But it does not. Thus, A is not 
less than B. And it was shown not (to be) equal either. 
Thus, A is greater than B. 

So, again, let C have a greater ratio to B than C (has) 
to A. I say that B is less than A. 

For if not, (it is) surely either equal or greater. In fact, 
B is not equal to A. For (then) C would have the same 
ratio to each of A and B [Prop. 5.7]. But it does not. 
Thus, A is not equal to B. Neither, indeed, is B greater 
than A. For (then) C would have a lesser ratio to B than 
(it has) to A [Prop. 5.8]. But it does not. Thus, B is not 
greater than A. And it was shown that (it is) not equal 
(to A) either. Thus, B is less than A. 

Thus, for (magnitudes) having a ratio to the same 
(magnitude), that (magnitude which) has the greater 
ratio is (the) greater. And that (magnitude) to which 
the latter (magnitude) has a greater ratio is (the) lesser. 
(Which is) the very thing it was required to show. 

Proposition lit 

(Ratios which are) the same with the same ratio are 
also the same with one another. 

A 1 C' 1 Ei 1 

B — i D 1 F' 1 

Gi 1 1 Hi 1 1 Ki 1 1 

Li — i — i — i M 1 1 1 N 1 1 1 1 

For let it be that as A (is) to B, so C (is) to D, and as 

C (is) to D, so E (is) to F. I say that as A is to B, so E 

(is) to F. 

For let the equal multiples G, H, K have been taken 
of A, C, E (respectively), and the other random equal 
multiples L, M, N of B, D, F (respectively). 

And since as A is to B, so C (is) to D, and the equal 
multiples G and H have been taken of A and C (respec- 
tively), and the other random equal multiples L and M 
of B and D (respectively), thus if G exceeds L then H 
also exceeds M, and if (G is) equal (to L then H is also) 



141 



ETOLXEIftN z. 



ELEMENTS BOOK 5 



(be; to r Ttpoc; to A, outcoc; to E Ttpoc; to Z, xdi dXrjUTai 
tcov r, E loaxic, TtoXXaTtXdaia Ta 0, K, tcov 5e A, Z dXXa, 
a etu^ev, iadxic; TtoXXaTtXdaia Ta M, N, el dpa UTtepexei TO 
9 tou M, uitepexei xal to K tou N, xal si laov, l'aov, xal el 
eXXaTov, eXaTTov. dXXa ei UTtepeTxe to O tou M, UTtepeTxe 
xal to H tou A, xal ei I'aov, I'aov, xal ei eXaTTov, eXaTTov 
cbaTe xal ei UTtepexei to H tou A, UTtepexei xal to K tou 
N, xal ei laov, I'aov, xal ei eXaTTov, eXaTTov. xai coti Ta 
uev H, K tcov A, E iadxic; TtoXXaTtXdaia, Ta Be A, N iSv B, 
Z dXXa, a CTU)(ev, iadxic; TtoXXaTtXdaia' cotiv dpa cbc; to A 
Ttpoc; to B, outcoc; to E Ttpoc; to Z. 

Oi dpa tco aincp Xoycp oi auToi xal dXXr|Xoic; eiaiv oi 
auToi- oTtep e8ei SeT^ai. 



equal (to M), and if (G is) less (than L then H is also) 
less (than M) [Def. 5.5]. Again, since as G is to D, so 
E (is) to F, and the equal multiples H and K have been 
taken of G and E (respectively), and the other random 
equal multiples M and N of D and F (respectively), thus 
if H exceeds M then K also exceeds N, and if (H is) 
equal (to M then K is also) equal (to AO, and if (H is) 
less (than M then K is also) less (than N) [Def. 5.5]. But 
(we saw that) if H was exceeding M then G was also ex- 
ceeding L, and if (H was) equal (to M then G was also) 
equal (to L), and if (H was) less (than M then G was 
also) less (than L). And, hence, if G exceeds L then K 
also exceeds AT, and if (G is) equal (to L then AT is also) 
equal (to AO, and if (G is) less (than L then K is also) 
less (than N). And G and K are equal multiples of A 
and (respectively), and L and A^ other random equal 
multiples of B and F (respectively). Thus, as A is to B, 
so E (is) to F [Def. 5.5]. 

Thus, (ratios which are) the same with the same ratio 
are also the same with one another. (Which is) the very 
thing it was required to show. 



t In modern notation, this proposition reads that if a : (3 :: 7 : 8 and 7 : S :: e : C then a : /? :: e : C- 



'Eav fj oTiooaouv [ieye'Or) dvaXoyov, eoTai cbc; ev tGv 
f)youuevcov npbc, ev tGv enotievwv, outgk anavTa Ta 
fjyouueva Ttpoc; aTtavTa Ta CTto^ieva. 

Ai ' r> ' ei — ' 

B' 1 ' Z' 1 

H 

©1 1 Mi 1 

K' ' N' ' 

'EaTwaav oTtoaaouv [icye'dr) dvaXoyov Ta A, B, T, A, 
E, Z, cbc; to A Ttpoc to B, outcoc; to T Ttpoc; to A, xai to E 
Ttpoc; to Z- Xeyw, oti cotIv wc; to A Ttpoc; to B, outioc; Ta 
A, r, E Ttpoc; Ta B, A, Z. 

EiXiqcp-dw yap tGv \iev A, F, E iadxic; TtoXXaTtXdaia Ta H, 
0, K, t£5v 8e B, A, Z dXXa, a ctu^cv, iadxic; TtoXXaTtXdaia 
Td A, M, N. 

Kai CTtei eaTiv cbc; to A Ttpoc; to B, outcoc; to T Ttpoc; to 
A, xal to E Ttpoc; to Z, xal el'XrjTtTai tcov (lev A, T, E iadxic; 
TtoXXaTtXdaia Ta H, Q, K tcov 5e B, A, Z dXXa, d eTU/ev, 
iadxic; TtoXXaTtXdaia Ta A, M, N, ei dpa UTtepexei to H tou A, 
UTtepexei xai to 8 tou M, xal to K tou N, xal ei laov, i'aov, 
xal ei eXaTTov, eXaTTov. cootc xai ei UTtepexei to H tou A, 



Proposition 12+ 

If there are any number of magnitudes whatsoever 
(which are) proportional then as one of the leading (mag- 
nitudes is) to one of the following, so will all of the lead- 
ing (magnitudes) be to all of the following. 

A' 1 C' ' E' ' 

B 1 D 1 F 1 



Hi 1 Mi 1 

Ki 1 Ni 1 

Let there be any number of magnitudes whatsoever, 
A, B, C, D, E, F, (which are) proportional, (so that) as 
A (is) to B, so G (is) to D, and E to F. I say that as A is 
to B, so A, C, E (are) to B, D, F. 

For let the equal multiples G, H, K have been taken 
of A, C, E (respectively), and the other random equal 
multiples L, M, N of B, D, F (respectively). 

And since as A is to B, so G (is) to D, and E to F, and 
the equal multiples G, H, K have been taken of A, C, E 
(respectively), and the other random equal multiples L, 
M, N of B, D, F (respectively), thus if G exceeds L then 
H also exceeds M, and K (exceeds) N, and if (G is) 
equal (to L then H is also) equal (to M, and K to N), 



A 1 G' 1 Li 



142 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



UTtepexei xod xd H, 0, K iwv A, M, N, xal ei iaov, laa, xal 
ei eXaxxov, eXaxxova. xai eaxi xo \itv H xal xd H, 0, K 
xoO A xal xfiv A, T, E ladxic TtoXXanXdaia, ETX£i8r]TXEp edv 
fj oTioaaouv [icyc'dr) oTtoawvouv [leycdfiv iowv xo TtXrj'doc 
exaaxov exdaxou ladxic TioXXanXdaiov, oaaitXdaiov eaxiv 
ev iSv ^teye-duv evoc, xoaauxauXdaia eaxai xal xd Ttdvxa 
iSv Tidvxcov. Bid xd auxd 8f] xal xo A xal xd A, M, N xoO 
B xal xfiv B, A, Z ladxu; eaxi TtoXXaitXdaia- eaxiv apa (be 
xo A npbc, xo B, ouxw<; xd A, T, E Tipoc xd B, A, Z. 

'Edv apa fj oTtoaaouv [icyc'dr] dvdXoyov, eaxai 6c ev 
xwv f]you^iv«v npbz ev xwv enofievuv, ouxgk auavxa xd 
f]yo6tX£va 7tp6<; arcavxa xd en6\ie\)&- ojiep eBei BeT^ai. 



and if (G is) less (than L then ii is also) less (than M, 
and K than AO [Def. 5.5]. And, hence, if G exceeds L 
then G, if, K also exceed L, M, N, and if (G is) equal 
(to L then G, ff, K are also) equal (to L, M, N) and 
if (G is) less (than L then G, if, if are also) less (than 
L, M, N). And G and G, if, if are equal multiples of 
A and A, G, f? (respectively), inasmuch as if there are 
any number of magnitudes whatsoever (which are) equal 
multiples, respectively, of some (other) magnitudes, of 
equal number (to them), then as many times as one of the 
(first) magnitudes is (divisible) by one (of the second), 
so many times will all (of the first magnitudes) also (be 
divisible) by all (of the second) [Prop. 5.1]. So, for the 
same (reasons), L and L, M, N are also equal multiples 
of B and B, D, F (respectively). Thus, as A is to B, so 
A, C, E (are) to B, D, F (respectively). 

Thus, if there are any number of magnitudes whatso- 
ever (which are) proportional then as one of the leading 
(magnitudes is) to one of the following, so will all of the 
leading (magnitudes) be to all of the following. (Which 
is) the very thing it was required to show. 



+ In modern notation, this proposition reads that if a : a' :: f} :/?':: 7 : 7' etc. then a : a' :: (a + + 7 H ) : (a' + (3' + 7' H ). 



'Edv upGxov npbz 8euxepov xov auxov !)(T) Xoyov xal 
xpixov 7tpo<; xexapxov, xplxov Be 7ip6<; xexapxov ^.sl^ova 
Xoyov exTi fj Tisjinxov npbz Ixxov, xal upaixov upoc; Beuxepov 
(lei^ova Xoyov s^ei fj ttejitixov npbz exxov. 

A ' r ' Ei ' 

B — ' A' ' Z ' 

Mi 1 iH' ' © ' ' 

N 1 — 1 — 1 — 1 K 1 1 1 1 A 1 1 1 1 

ripGxov yap xo A Tipoc; Beuxepov xo B xov auxov e)(exw 
Xoyov xal xpixov xo T npbq xfxapxov xo A, xpixov Be xo T 
7ip6<; xexapxov xo A ^ie[£ova Xoyov exexw r\ TieuTixov xo E 
Tipoc; exxov xo Z. Xeyco, 0x1. xal Tipoxov xo A 7ip6<; Beuxepov 
xo B [iciCova Xoyov e^ei fjuep TtejiTtxov xo E Tipoc; exxov xo 
Z. 

'EticI yap eaxi xivd xov ^xev T, E ladxic TioXXauXdaia, 
xCSv Be A, Z dXXa, a exu)(ev, ladxic noXXaTiXdoia, xal xo (iev 
xoO r TioXXaTiXdoiov xou xou A TtoXXauXaabu bnspiyzi, 
xo Be xou E TioXXajcXdaiov xou xou Z TioXXaicXaaiou ou)( 
ujcepexei, EiXfjqj'dco, xal eaxw xwv ^.ev T, E ladxic tcoX- 
XaicXdaia xd H, 0, xwv Be A, Z dXXa, a exu)(ev, ladxic 
TioXXauXdaia xd K, A, «axe xo [lev H xou K U7iepe)(eiv, xo 
Be xou A \±r\ uuepexeiv xal oaaTiXdaiov (lev eaxi xo H 
xou T, xoaauxanXdaiov eaxco xal xo M xou A, oaauXdaiov 
Be xo K xou A, xoaauxanXdaiov eaxw xal xo N xou B. 



Proposition 13 f 

If a first (magnitude) has the same ratio to a second 
that a third (has) to a fourth, and the third (magnitude) 
has a greater ratio to the fourth than a fifth (has) to a 
sixth, then the first (magnitude) will also have a greater 
ratio to the second than the fifth (has) to the sixth. 

A' ' C ' E' 1 

B 1 1 D 1 F' 1 

Mi 1 1 G 1 ' Hi 1 1 

Ni 1 1 1 K 1 1 1 Li 1 1 1 

For let a first (magnitude) A have the same ratio to a 
second B that a third G (has) to a fourth D, and let the 
third (magnitude) G have a greater ratio to the fourth 
D than a fifth E (has) to a sixth F. I say that the first 
(magnitude) A will also have a greater ratio to the second 
B than the fifth E (has) to the sixth F. 

For since there are some equal multiples of G and 
E, and other random equal multiples of D and F, (for 
which) the multiple of G exceeds the (multiple) of D, 
and the multiple of E does not exceed the multiple of F 
[Def. 5.7], let them have been taken. And let G and H be 
equal multiples of G and E (respectively), and K and L 
other random equal multiples of D and F (respectively), 
such that G exceeds K, but H does not exceed L. And as 
many times as G is (divisible) by G, so many times let M 
be (divisible) by A. And as many times as K (is divisible) 



143 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



Kal ETtei eaxiv w<; to A Ttpoc; to B, ouxco? to Y Ttpoc; 
to A, xal eiXrjTtxai xfiiv jjiev A, Y iadxic; TtoXXaTtXdaia xd 
M, H, iSv 8e B, A aXXa, a exuxev, iadxic; TtoXXaTtXdaia xd 
N, K, el dpa UTteps/si to M xou N, UTtepexei xal xo H xou 
K, xod el laov, Taov, xal ei eXaxxov, eXXaxov. Cmepexei Be 
xo H xou K- UTtepexei dpa xal xo M xou N. xo Se 6 xou 
A oux UTtepexei 1 xai eaxi xd [iev M, O xaiv A, E iadxic; 
TtoXXaTtXdaia, xd 6e N, A xfiv B, Z aXXa, a exuxev, Iadxic; 
TtoXXaTtXdaia- xo dpa A Ttpoc; xo B ^tei^ova Xoyov exei f]nep 
xo E Ttpoc; xo Z. 

'Eav dpa upwxov Ttpoc; Beuxepov xov auxov ex?] Xoyov 
xal xpixov Ttpoc; xexapxov, xpixov 8e Ttpoc; xexapxov ^tei^ova 
Xoyov exT) ^ Tte^tTtxov Ttpoc; exxov, xal TtpGxov Ttpoc; 8euxepov 
^tei^ova Xoyov ei;ei rj Tte^titxov Ttpoc; exxov oitep e8ei SeTc;ai. 



by D, so many times let N be (divisible) by B. 

And since as ^4 is to B, so C (is) to D, and the equal 
multiples M and G have been taken of A and G (respec- 
tively), and the other random equal multiples N and K 
of B and D (respectively), thus if M exceeds N then G 
exceeds K, and if (M is) equal (to N then G is also) 
equal (to K), and if (M is) less (than N then G is also) 
less (than K) [Def. 5.5]. And G exceeds K. Thus, M 
also exceeds N. And i7 does not exceeds L. And M and 
H are equal multiples of A and _E (respectively), and N 
and L other random equal multiples of B and F (respec- 
tively). Thus, A has a greater ratio to B than (has) to 
F [Def. 5.7]. 

Thus, if a first (magnitude) has the same ratio to a 
second that a third (has) to a fourth, and a third (magni- 
tude) has a greater ratio to a fourth than a fifth (has) 
to a sixth, then the first (magnitude) will also have a 
greater ratio to the second than the fifth (has) to the 
sixth. (Which is) the very thing it was required to show. 



t In modern notation, this proposition reads that if a : (3 :: 7 : 8 and 7 : 8 > e : f then a : (3 > e : (. 



i5'. 

'Eav TtpcSxov Ttpoc; Seuxepov xov auxov exT) Xoyov xal 
xpixov Ttpoc; xexapxov, xo Se TtpGxov xou xpixou ^icTCov fj, 
xal xo Seuxepov xou xexdpxou jieT^ov eaxai, xdv Taov, '(gov, 
xdv eXaxxov, eXaxxov. 

Ai 1 ri 1 

Bi 1 A' ' 

IlpGxov yap TO A Ttpoc; Seuxepov xo B auxov cxcxm 
Xoyov xal xpixov xo Y Ttpoc; xexapxov xo A, [leTC^ov Se eaxw 
xo A xou r- Xeyw, oxi xal xo B xou A fjieT^ov eaxiv. 

'Eitel yap xo A xou Y \j.e%6\i eaxiv, dXXo Se, o exuxev, 
[^eyedoc;] xo B, xo A dpa Ttpoc; xo B ^tei^ova Xoyov exei 
rjitep xo T Ttpoc; xo B. (be; Se xo A Ttpoc; xo B, ouxwc; xo 
T Ttpoc; xo A- xal xo T dpa Ttpoc xo A ^eii^ova Xoyov exei 
f]Ttep xo r Ttpoc; xo B. Ttpoc; o Se xo auxo (jiei^ova Xoyov 
exei, exelvo eXaaaov eaxiv eXaaaov dpa xo A xou B- waxe 
^el£6v eaxi xo B xou A. 

"O^oicoc; 8rj 8eTi;o^ev, 6x1 xdv taov f) xo A xw Y, Taov 
eaxai xal xo B iu A, xdv eXaaaov fj xo A xou T, eXaaaov 
eaxai xal xo B xou A. 

'Eav dpa itpwxov Ttpoc; Seuxepov xov auxov exT) Xoyov 
xal xpixov Ttpoc; xexapxov, xo 8e Ttpwxov xou xpixou y.e%o\> fj, 
xal xo 8euxepov xou xexdpxou \ie%o\> eaxai, xdv Taov, Taov, 
xdv eXaxxov, eXaxxov oitep e8ei 8ele;ai. 



Proposition 14 f 

If a first (magnitude) has the same ratio to a second 
that a third (has) to a fourth, and the first (magnitude) 
is greater than the third, then the second will also be 
greater than the fourth. And if (the first magnitude is) 
equal (to the third then the second will also be) equal (to 
the fourth) . And if (the first magnitude is) less (than the 
third then the second will also be) less (than the fourth) . 

A 1 1 C' 1 

B 1 1 D 1 

For let a first (magnitude) A have the same ratio to a 
second B that a third G (has) to a fourth D. And let A be 
greater than G. I say that B is also greater than D. 

For since A is greater than G, and B (is) another ran- 
dom [magnitude], A thus has a greater ratio to B than G 
(has) to B [Prop. 5.8]. And as A (is) to B, so G (is) to 
D. Thus, G also has a greater ratio to D than G (has) to 
B. And that (magnitude) to which the same (magnitude) 
has a greater ratio is the lesser [Prop. 5.10]. Thus, D (is) 
less than B. Hence, B is greater than D. 

So, similarly, we can show that even if A is equal to G 
then B will also be equal to D, and even if A is less than 
G then B will also be less than D. 

Thus, if a first (magnitude) has the same ratio to a 
second that a third (has) to a fourth, and the first (mag- 
nitude) is greater than the third, then the second will also 
be greater than the fourth. And if (the first magnitude is) 



144 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



equal (to the third then the second will also be) equal (to 
the fourth) . And if (the first magnitude is) less (than the 
third then the second will also be) less (than the fourth) . 
(Which is) the very thing it was required to show. 

t In modern notation, this proposition reads that if a : j3 :: 7 : 8 then a = 7 as f3 = <5. 



IE . 



Proposition 15 f 



Td H£pT] xolc; Gaauxioc; TtoXXauXaaioic; xov aCxov ttyzi 
Xoyov XTjcpiSevxa xaxdXXr]Xa. 

A H B 

1 1 1 1 j/ 1 1 1 



A 

1— 



K 

— 1 — 



A 

— 1 — 



E 

—1 



2' ' 

"Eaxw yap ladxic; TtoXXauXdaiov to AB xou T xal to AE 
toO Z- Xey«, oxi laxlv cbc xo T Ttpoc; xo Z, ouxwc; xo AB 
7ip6<; xo AE. 

'End yap ladxic; eaxl KoXXairXdoiov xo AB xoO T xal 
xo AE xou Z, 00a apa eaxlv ev xG AB ^.eyei!)/] laa xG 
T, xoaaOxa xal ev xG AE Taa xG Z. 5ir]pr]OTf)w xo fjiev AB 
slg xa xG T I'oa xd AH, H6, 6B, xo 8e AE zlc. xd xG Z 
Taa xa AK, KA, AE' eaxai 8r| I'aov xo TtXrj'doc; xGv AH, 
H6, 6B xG TrXrydei xGv AK, KA, AE. xal end Taa eoxl xa 
AH, H6, 9B dXX^Xoic, eoxi Bs xal xd AK, KA, AE I'oa 
dXXiqXou;, saxiv apa Gc; xo AH npbc, xo AK, ouxwc; xo H0 
Tipog xo KA, xal xo &B npbz xo AE. soxai apa xal Gg §v 
xGv f)you|jievMv Ttpoc; ev xGv euo^ievwv, ouxwc drcavxa xd 
f)you[ieva upoc duavxa xd snojieva' eaxiv apa G<; xo AH 
npbc, xo AK, ouxgk xo AB npbc, xo AE. I'oov 8e xo jisv AH 
xG T, xo 8s AK xG Z- eoxiv apa Gc; xo T npbc, xo Z ouxcoc 
xo AB iipoc; xo AE. 

Td apa [jtspr] xoTg Gaauxwc; TioXXanXaoioic; xov auxov 
SX^I Xoyov XricpiSevxa xaxdXXrjXa- oiiep eBei BeT^ai. 



Parts have the same ratio as similar multiples, taken 
in corresponding order. 

A G H B 

1 1 1 1 Q 1 1 



D 

1— 



K 



E 

—1 



For let AB and DE be equal multiples of C and F 
(respectively) . I say that as C is to F, so AB (is) to DE. 

For since AB and DE are equal multiples of C and 
F (respectively), thus as many magnitudes as there are 
in AB equal to C, so many (are there) also in DE equal 
to F. Let AB have been divided into (magnitudes) AG, 
GH, HB, equal to C, and DE into (magnitudes) DK, 
KL, LE, equal to F. So, the number of (magnitudes) 
AG, GH, HB will equal the number of (magnitudes) 
DK, KL, LE. And since AG, GH, HB are equal to one 
another, and DK, KL, LE are also equal to one another, 
thus as AG is to DK, so GH (is) to KL, and HB to LE 
[Prop. 5.7]. And, thus (for proportional magnitudes), as 
one of the leading (magnitudes) will be to one of the fol- 
lowing, so all of the leading (magnitudes will be) to all of 
the following [Prop. 5.12]. Thus, as AG is to DK, so AB 
(is) to DE. And AG is equal to C, and DK to F. Thus, 
as C is to F, so AB (is) to DE. 

Thus, parts have the same ratio as similar multiples, 
taken in corresponding order. (Which is) the very thing 
it was required to show. 



t In modern notation, this proposition reads that a : f3 : : m a : m f3. 



If'. 

'Edv xeoaapa (ieyei9r) dvdXoyov f), xal evaXXac; dvdXoyov 
eaxai. 

"Eax« xeaoapa [leye^r] dvdXoyov xd A, B, T, A, G<; xo 
A Tipoc xo B, ouxox xo T Tipoc; xo A- Xcyw, oxi xal svaXXd^ 
[dvdXoyov] eaxai, G<; xo A npoc; xo T, ouxwc; xo B Tipoc xo 
A. 

ElXr ! ]cp'dw yap xGv jiev A, B ladxic; TtoXXaTtXaaia xd E, 
Z, xGv 8e T, A aXXa, a sxuxev, ladxic; TioXXajcXdaia xd H, 

e. 



Proposition 16 f 

If four magnitudes are proportional then they will also 
be proportional alternately. 

Let A, B, C and D be four proportional magnitudes, 
(such that) as A (is) to B, so C (is) to D. I say that they 
will also be [proportional] alternately, (so that) as A (is) 
to C, so B (is) to D. 

For let the equal multiples E and F have been taken 
of A and B (respectively), and the other random equal 
multiples G and H of C and D (respectively) . 



145 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



B h 

E i- 
Z^ 



H 1 



Kal cticI iadxic eaxl TtoXXauXdaiov to E xoO A xal to Z 
toO B, xa Se (JepT) xolc waauxwc TtoXXaTiXaaioic xov aGxov 
exei Xoyov, eaxiv apa cbc xo A Tipoc xo B, ouxwc xo E Tipoc 
xo Z. «c 8e xo A Tipoc xo B, ouxwc xo T Tipoc xo A' xal d>c 
apa xo r Tipoc xo A, ouxwc xo E Tipoc xo Z. TtdXiv, euel xd 
H, 6 xwv T, A iadxic eaxl TioXXaTtXdaia, eaxiv apa cbc xo T 
Tipoc xo A, ouxoc xo H Tipoc xo 0. ci>c 8e xo T Tipoc xo A, 
[ouxtoc] xo E Tipoc xo Z- xal <i>c apa xo E Tipoc xo Z, ouxcoc 
xo H Tipoc xo 9. eav 8e xeaaapa ^eyc'dr] dvdXoyov fj, xo 8s 
Ttpwxov xoO xpixou [isT^ov fj, xal xo 8euxepov xou xexdpxou 
\ieiZov eaxai, xav laov, iaov, xdv eXaxxov, eXaxxov. ei apa 
UTiepexei xo E xou H, imepexei xal xo Z xou 0, xal ei laov, 
laov, xal ei eXaxxov, eXaxxov. xal eaxi xd ^.ev E, Z xwv 
A, B iadxic TioXXaTtXdaia, xa 8e H, 8 xov T, A aXXa, a 
exuxev, iadxic TioXXaTtXdaia- eaxiv apa cbc xo A Tipoc xo T, 
ouxwc xo B Tipoc xo A. 

'Eav apa xeaaapa ^.eyedr] dvdXoyov fj, xal evaXXa^ 
dvdXoyov eaxar oxep e8ei SeT^ai. 



A^ 
F 



O 



H 1 



And since E and F are equal multiples of A and B 
(respectively), and parts have the same ratio as similar 
multiples [Prop. 5.15], thus as A is to B, so E (is) to F. 
But as A (is) to B, so C (is) to D. And, thus, as C (is) 
to D, so _E (is) to F [Prop. 5.11]. Again, since G and H 
are equal multiples of C and D (respectively), thus as C 
is to D, so G (is) to Jf [Prop. 5.15]. But as C (is) to D, 
[so] £ (is) to F. And, thus, as E (is) to F, so G (is) to 
iJ [Prop. 5.11]. And if four magnitudes are proportional, 
and the first is greater than the third then the second will 
also be greater than the fourth, and if (the first is) equal 
(to the third then the second will also be) equal (to the 
fourth), and if (the first is) less (than the third then the 
second will also be) less (than the fourth) [Prop. 5.14]. 
Thus, if E exceeds G then F also exceeds H, and if (E is) 
equal (to G then F is also) equal (to H), and if (E is) less 
(than G then F is also) less (than H) . And E and F are 
equal multiples of A and B (respectively), and G and H 
other random equal multiples of C and D (respectively) . 
Thus, as A is to C, so B (is) to D [Def. 5.5]. 

Thus, if four magnitudes are proportional then they 
will also be proportional alternately. (Which is) the very 
thing it was required to show. 



t In modern notation, this proposition reads that if a : (3 :: 7 : 8 then a : 7 :: (3 : 8. 



'Edv auyxei^ieva jieye-v}/] dvdXoyov fj, xal 8iaipe-devxa 
dvdXoyov eaxai. 

A E B r Z A 

1 1 1 1 1 — 1 

H S K S 

1 1 1 1 

A M N n 

1 1 1 1 

TCaxco auyxeifieva [ieye'Or] dvdXoyov xa AB, BE, IA, 
AZ, cbc to AB Tipoc xo BE, ouxwc to IA Tipoc to AZ- 
Xeyco, oxi xal Biaipcdevxa dvdXoyov eaxai, wc xo AE Tipoc 
xo EB, ouxwc to TZ Tipoc to AZ. 

EiXrjcp'dco yap xwv ^tev AE, EB, TZ, ZA iadxic TioX- 
XaTtXdaia xa H9, 6K, AM, MN, xwv 8e EB, ZA aXXa, a 
exuxev, iadxic TioXXaTtXdaia xa KS, Nil. 

Kal cticI iadxic saxl TioXXaxXdaiov xo H9 xou AE xal 
xo OK xou EB, iadxic apa eaxl TtoXXaTiXdaiov xo H6 xoO 



Proposition 17 f 

If composed magnitudes are proportional then they 
will also be proportional (when) separarted. 

A E B C F D 

1 1 1 1 1 1 

G H K O 

1 1 1 1 

L MNP 

1 1 1 1 

Let AB, BE, CD, and DF be composed magnitudes 
(which are) proportional, (so that) as AB (is) to BE, so 
CD (is) to DF. I say that they will also be proportional 
(when) separated, (so that) as AE (is) to EB, so CF (is) 
to DF. 

For let the equal multiples GH, HK, LM, and MN 
have been taken of AE, EB, CF, and FD (respectively), 
and the other random equal multiples KO and NP of 
EB and FD (respectively). 



146 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



AE xal to HK tou AB. iadxu; Se ecru TtoXXaitXaaiov to H0 
tou AE xal to AM tou TZ- iodxn; apa eaTi TtoXXanXdmov 
to HK tou AB xal to AM toO TZ. TtdXiv, snel iadxic sgtI 
TtoXXarcXaaiov to AM tou TZ xal to MN tou ZA, iadxu; apa 
taxi TroXXanXdaiov to AM tou TZ xal to AN tou IA. ladxig 
Be rjv KoXXanXdoiov to AM tou TZ xal to HK tou AB- 
iadxu; apa eaTi TioXXanXdoiov to HK tou AB xal to AN tou 
TA. Ta HK, AN apa t«v AB, IA iadxu; eotI TtoXXanXdaia. 
ndXiv, inz\ iadxu; eotI TioXXaTtXaaiov to 0K tou EB xal to 
MN tou ZA, eaTi Be xal to KS tou EB iadxu; TioXXanXdaiov 
xal to Nn tou ZA, xal auvTcdev to 9S tou EB iadxu; eaxl 
TtoXXarcXaaiov xal to Mil tou ZA. xal snei eaTiv cbc to AB 
7ip6<; to BE, outo<; to TA 7ip6<; to AZ, xal d'Xr]7iTai xwv 
[lev AB, TA iadxu; TtoXXaTtXdaia Ta HK, AN, t£Sv 8e EB, 
ZA iadxu; TtoXXaTtXdaia Ta 9S, Mil, ei apa (mepexei TO 
HK tou 9S, unepexei xal to AN tou Mil, xal ei laov, laov, 
xal ei sXaTTov, eXaTTOv. uitepexeTO) 8f| t° HK tou OS, 
xal xoivou dcpaipcdevTOi; tou 9K UTiepexei. apa xal to H6 
tou KS. dXXa ei U7iepeT)(e to HK tou 0S uiiepeTxe xal to 
AN tou Mil' UTiepexei apa xal to AN tou Mil, xal xoivou 
dcpaipcdevTot; tou MN UTiepexei xal to AM tou Nn- wcrce 
ei UTiepexei to H8 tou KS, uuepexei xal to AM tou Nn. 
o^ioicoc; 8r] SeT^o^tev, oxi xdv laov fj to H8 to KS, taov 
eaTai xal to AM to Nn, xdv eXaTTOv, eXaTTOv. xai eaTi Ta 
[lev H0, AM twv AE, TZ iadxu; TioXXaicXdaia, Ta 8e KS, 
Nn twv EB, ZA dXXa, a ctuxcv, iadxic; icoXXaiiXdaia- cotiv 
apa ok to AE 7ip6<; to EB, outoc to TZ Ttpoc; to ZA. 

'Eav apa auyxei^ieva [leye'dr) dvdXoyov fj, xal 8iai- 
peiDevTa dvdXoyov eaTar oiiep e8ei SeT^ai. 



And since GH and HK are equal multiples of AE and 
EB (respectively), GH and GK are thus equal multiples 
of AE and AB (respectively) [Prop. 5.1]. But GH and 
LM are equal multiples of AE and CF (respectively). 
Thus, GK and LM are equal multiples of AB and CF 
(respectively) . Again, since LM and MN are equal mul- 
tiples of CF and FD (respectively), LM and LN are thus 
equal multiples of CF and CD (respectively) [Prop. 5.1]. 
And LM and GK were equal multiples of CF and AB 
(respectively). Thus, GK and LN are equal multiples 
of AB and CD (respectively). Thus, GK, LN are equal 
multiples of AB, CD. Again, since HK and MN are 
equal multiples of EB and FD (respectively), and KO 
and NP are also equal multiples of EB and FD (respec- 
tively), then, added together, HO and MP are also equal 
multiples of FP> and FD (respectively) [Prop. 5.2]. And 
since as AB (is) to BE, so CD (is) to DF, and the equal 
multiples GK, LN have been taken of AB, CD, and the 
equal multiples HO, MP of EB, FD, thus if GK exceeds 
HO then LN also exceeds MP, and if {GK is) equal (to 
HO then LN is also) equal (to MP), and if (GK is) less 
(than HO then LN is also) less (than MP) [Def. 5.5]. 
So let GK exceed HO, and thus, HK being taken away 
from both, GH exceeds KO. But (we saw that) if GK 
was exceeding HO then LN was also exceeding MP. 
Thus, LN also exceeds MP, and, MN being taken away 
from both, LM also exceeds NP. Hence, if GH exceeds 
KO then LM also exceeds NP. So, similarly, we can 
show that even if GH is equal to KO then LM will also 
be equal to NP, and even if (GH is) less (than KO then 
LM will also be) less (than NP). And GH, LM are equal 
multiples of AE, CF, and KO, NP other random equal 
multiples of EB, FD. Thus, as AE is to EB, so CF (is) 
to FD [Def. 5.5]. 

Thus, if composed magnitudes are proportional then 
they will also be proportional (when) separarted. (Which 
is) the very thing it was required to show. 



t In modern notation, this proposition reads that ifa + /3:/3::7 + <5:<5 then a : f3 :: 7 : 8. 



'Edv 8ir)p/)[ieva ^eye^r) dvdXoyov fj, xal auvTeiJevTa 
dvdXoyov eaxai. 

A E B 



Proposition 18+ 

If separated magnitudes are proportional then they 
will also be proportional (when) composed. 

A E B 



r 

i— 



z 

— i — 



H 

—i — 



A 

— i 



'Eotw 8ir]pr]neva [icye-dr] dvdXoyov Ta AE, EB, TZ, ZA, 
&>c, to AE Ttpoc; to EB, outoc; to TZ Ttpoc; to ZA- Xeyw, 
oti xal ouvTeiJevTa dvdXoyov eaxai, tbc; to AB npbc, to BE, 



c 

h- 



G 

— i — 



D 

— i 



Let AE, EB, CF, and FD be separated magnitudes 
(which are) proportional, (so that) as AE (is) to EB, so 
CF (is) to FD. I say that they will also be proportional 



147 



ETOIXEIftN z. 



ELEMENTS BOOK 5 



ouxcoc; to TA Ttpoc; to ZA. 

EE yap \xr\ eaxiv tlx; to AB Ttpoc; xo BE, ouxcoc; to TA 
Ttpoc; xo AZ, eaxai cbc; xo AB Ttpoc; xo BE, ouxcoc; xo TA 
fjxoi Ttpoc; eXaaaov xi xou AZ fj Ttpoc; jiel^ov. 

'Eaxco Ttpoxepov Ttpoc; eXaaaov xo AH. xal stxei eaxiv cbc; 
xo AB Ttpoc; xo BE, ouxcoc; xo TA Ttpoc; xo AH, auyxei^eva 
(leyc'dr) dvaXoyov eaxiv wore xal Biaipcdevxa dvaXoyov 
eaxai. eaxiv dpa cbc; xo AE Ttpoc; xo EB, ouxcoc; xo TH Ttpoc; 
xo HA. UTtoxeixai 5e xal cbc; xo AE Ttpoc; xo EB, ouxcoc; xo 
rZ Ttpoc; xo ZA. xal cbc; dpa xo TH Ttpoc; xo HA, ouxcoc; xo 
rZ Ttpoc; xo ZA. jiel^ov 6e xo Ttpcoxov xo TH xou xpixou xou 
TZ- [iel^ov dpa xal xo 8euxepov xo HA xou xexdpxou xou 
ZA. dXXd xal eXaxxov oTtep eaxiv dBuvaxov oux dpa eaxiv 
cbc; xo AB Ttpoc; xo BE, ouxcoc; xo TA Ttpoc; eXaaaov xou 
ZA. ojioloK 6f| 6eii;o^ev, oxi ouSe Ttpoc; [MCov Ttpoc; auxo 
dpa. 

'Eav dpa Biflprjjieva |ieyei9r) dvaXoyov rj, xal auvxe-devxa 
dvaXoyov eaxai- onep eSei 8el^ai. 



(when) composed, (so that) as AB (is) to BE, so CD (is) 
to FD. 

For if (it is) not (the case that) as AB is to BE, so 
CD (is) to FD, then it will surely be (the case that) as 
AB (is) to BE, so CD is either to some (magnitude) less 
than DF, or (some magnitude) greater (than DF)} 

Let it, first of all, be to (some magnitude) less (than 
DF), (namely) DC And since composed magnitudes 
are proportional, (so that) as AB is to BE, so CD (is) to 
DC, they will thus also be proportional (when) separated 
[Prop. 5.17]. Thus, as AE is to EB, so CG (is) to CD. 
But it was also assumed that as AE (is) to EB, so CF 
(is) to FD. Thus, (it is) also (the case that) as CG (is) 
to GD, so CF (is) to FD [Prop. 5.11]. And the first 
(magnitude) CG (is) greater than the third CF. Thus, 
the second (magnitude) GD (is) also greater than the 
fourth FD [Prop. 5.14]. But (it is) also less. The very 
thing is impossible. Thus, (it is) not (the case that) as AB 
is to BE, so CD (is) to less than FD. Similarly, we can 
show that neither (is it the case) to greater (than FD). 
Thus, (it is the case) to the same (as FD) . 

Thus, if separated magnitudes are proportional then 
they will also be proportional (when) composed. (Which 
is) the very thing it was required to show. 



t In modern notation, this proposition reads that if a : (3 :: 7 : <5 then a + /3 : /3 :: 7 + <5 : <5. 

* Here, Euclid assumes, without proof, that a fourth magnitude proportional to three given magnitudes can always be found. 



°Edv fj ok oXov Ttpoc oXov, ouxcoc; dcpaipsiJev Ttpoc; dcpai- 
pe^Ev, xal xo Xoikov Tcpoc; xo Xomov eaxai «<; oXov Tcpoc; 
oXov. 

A E B 

1 1 1 

r z a 

1 1 1 

'Eaxto yap tbcj oXov xo AB Ttpoc; oXov xo TA, ouxcoc; 
dcpaipe'dev xo AE Ttpoc; dcpEipeiDev xo TZ' Xeyco, oxi xal 
Xomov xo EB Ttpoc; Xomov xo ZA eaxai cbc; oXov xo AB 
Ttpoc; oXov xo TA. 

'EtccI yap eaxiv cbc; xo AB Ttpoc; xo TA, ouxcoc; xo AE 
Ttpoc; xo TZ, xal evaXXdc; cbc; xo BA Ttpoc; xo AE, ouxcoc; 
xo AT Ttpoc; xo TZ. xal enel auyxeijieva jieye-d/) dvaXoyov 
eaxiv, xal 8iaipei5evxa dvaXoyov eaxai, cbc; xo BE Ttpoc; xo 
EA, ouxcoc; xo AZ Ttpoc; xo TZ- xal evaXXdc;, cbc; xo BE Ttpoc; 
xo AZ, ouxcoc; xo EA Ttpoc; xo ZT. cb<; 8e xo AE Ttpoc; xo TZ, 
ouxcoc; UTtoxeixai oXov xo AB Ttpoc; oXov xo TA. xal XoiTtov 
dpa xo EB Ttpoc; Xoitiov xo ZA eaxai cbc; oXov xo AB Ttpog 
oXov xo TA. 

'Eav dpa fj cbc; oXov Ttpoc oXov, ouxcoc; dtpaipeiJev Ttpoc; 



Proposition 19 f 

If as the whole is to the whole so the (part) taken 
away is to the (part) taken away then the remainder to 
the remainder will also be as the whole (is) to the whole. 

A E B 

1 1 1 

C F D 

1 1 1 

For let the whole AB be to the whole CD as the (part) 
taken away AE (is) to the (part) taken away CF. I say 
that the remainder EB to the remainder FD will also be 
as the whole AB (is) to the whole CD. 

For since as AB is to CD, so AE (is) to CF, (it is) 
also (the case), alternately, (that) as BA (is) to AE, so 
DC (is) to CF [Prop. 5.16]. And since composed magni- 
tudes are proportional then they will also be proportional 
(when) separated, (so that) as BE (is) to EA, so DF (is) 
to CF [Prop. 5.17]. Also, alternately, as BE (is) to DF, 
so EA (is) to FC [Prop. 5.16]. And it was assumed that 
as AE (is) to CF, so the whole AB (is) to the whole CD. 
And, thus, as the remainder EB (is) to the remainder 
FD, so the whole AB will be to the whole CD. 



148 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



dcpaipsiftsv, xal to Xomov Ttpoc to Xoikov saTai <i>c oXov 
Ttpoc oXov [oTtsp s8si Ssl^ai]. 

[Kod STtel eSsix'dr) cbc to AB Ttpoc to IA, outwc to EB 
Ttpoc to ZA, xal svaXXdc" tlx; to AB Ttpoc to BE outgjc to 
TA Ttpoc to ZA, auyxsi|isva apa ^sysi}/] dvdXoyov screw 
sSeix'dr) 8s cbc to BA Ttpoc to AE, outioc to Ar Ttpoc to 
TZ- xai egtiv dvaoTps^avTi]. 



Thus, if as the whole is to the whole so the (part) 
taken away is to the (part) taken away then the remain- 
der to the remainder will also be as the whole (is) to 
the whole. [(Which is) the very thing it was required to 
show] 

[And since it was shown (that) as AB (is) to CD, so 
EB (is) to FD, (it is) also (the case), alternately, (that) 
as AB (is) to BE, so CD (is) to FD. Thus, composed 
magnitudes are proportional. And it was shown (that) 
as BA (is) to AE, so DC (is) to CF. And (the latter) is 
converted (from the former).] 



II6piG[jia. 

'Ex 8r] toutou cpavspov, oti sav auyxsijisva [isys'dr) 
dvdXoyov fj, xal dvacrcpsij'avTi dvdXoyov so"tar OTtsp s8si 
BsTc;ai. 

t In modern notation, this proposition reads that if a : (3 :: 7 : 5 then a : 
* In modern notation, this corollary reads that if a : (3 :: 7 : 8 then a : a 



Corollary 11 

So (it is) clear, from this, that if composed magni- 
tudes are proportional then they will also be proportional 
(when) converted. (Which is) the very thing it was re- 
quired to show. 

:: a - 7 : /3 - 8. 
— (3 :: 7 : 7 — 8. 



X . 

'Edv fj Tpia ^sys^r) xot ' dXXa auTolc Xai to TtXrji&oc, 
auv8uo Xa(iPav6(jieva xal sv tw ai)TO Xoyw, 81' I'aou Bs to 

IXpWTOV TOU TpiTOU [ISlCoV fj, Xal TO TSTapTOV TOU SXTOU 

[isiZov scrcai, xav I'oov, I'oov, xav sXaTTov, sXaTTov. 



A' 1 A 1 1 

Bi 1 Ei 1 

Ti 1 Z' 1 

'EaTW Tpia ^.sysiDr] Ta A, B, T, xal dXXa auTolc laa to 
TtXrj$oc Ta A, E, Z, auvBuo Xa^pavo^teva sv tw auTS Xoycp, 
cbc [isv to A Ttpoc to B, outwc to A Ttpoc to E, «c 8s to B 
Ttpoc to T, outmc to E Ttpoc to Z, 81' laou 8s fisT^ov sotw 
to A toO T- Xsyio, oti xal to A tou Z [isi^ov iaxai, xav 
laov, taov, xav sXaTTov, sXaTTov. 

'Etcei yap ^isT£6v scm to A toO T, aXXo 8s ti to B, to 8s 
^isT£ov Ttpoc to auTO [isiCova Xoyov sxsi fjusp to sXaTTov, 
to A apa Ttpoc to B (isi^ova Xoyov s^si fjTtsp to T Ttpoc to 
B. dXX' cbc ^isv to A Ttpoc to B [outoc] to A Ttpoc to E, cbc 
8s to T Ttpog to B, dvdnaXiv outgjc to Z Ttpoc to E- xal to 
A apa Ttpoc to E [isi^ova Xoyov s^si fjTtsp to Z Ttpoc to E. 
twv Bs Ttpoc to ai)TO Xoyov sxovtcjv to ^.s[£ova Xoyov s/ov 
[Lei^bv scmv. [leiZoM apa to A toO Z. o^tolwc 8r] 8s[^o[jisv, 
oti xav I'aov fj to A tw T, urov saTai xal to A t« Z, xav 



Proposition 20 f 

If there are three magnitudes, and others of equal 
number to them, (being) also in the same ratio taken two 
by two, and (if), via equality, the first is greater than the 
third then the fourth will also be greater than the sixth. 
And if (the first is) equal (to the third then the fourth 
will also be) equal (to the sixth). And if (the first is) less 
(than the third then the fourth will also be) less (than the 
sixth) . 

A 1 ' D' 1 

B 1 1 E 1 1 

Ci 1 F 1 1 

Let A, B, and C be three magnitudes, and D, E, F 
other (magnitudes) of equal number to them, (being) in 
the same ratio taken two by two, (so that) as A (is) to B, 
so D (is) to E, and as B (is) to C, so E (is) to F. And let 
A be greater than C, via equality. I say that D will also 
be greater than F. And if (A is) equal (to C then D will 
also be) equal (to F). And if (A is) less (than C then D 
will also be) less (than F). 

For since A is greater than C, and B some other (mag- 
nitude), and the greater (magnitude) has a greater ratio 
than the lesser to the same (magnitude) [Prop. 5.8], A 
thus has a greater ratio to B than C (has) to B. But as A 
(is) to B, [so] D (is) to E. And, inversely, as C (is) to B, 
so F (is) to E [Prop. 5.7 corr.]. Thus, D also has a greater 
ratio to E than F (has) to E [Prop. 5.13]. And for (mag- 



149 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



eXaxxov, eXaxxov. 

'Eav apa fj xpia (ieye'dr) xal aXXa auxolc; laa xo TtXrji&oc;, 
auvSuo Xa^pav6|ieva xal ev xw auxw Xoyw, 81° 1'aou 8e xo 
npwxov xou xpixou [isTCov fj, xal xo xexapxov xou exxou 
jieT£ov eaxai, xav iaov, iaov, xav eXaxxov, eXaxxov oTtep 
£8ei 8eTc;ai. 



t In modern notation, this proposition reads that if a : (3 :: 8 : e and j3 : 

XOL. 

'Eav fj xpia jieye'dr] xal aXXa auxolc; laa xo TtXrjT9oc; 
auv8uo Xajj.f3av6jj.sva xal ev xS auxfi Xoyw, fj 8e xexa- 
payjjevrj auxfiiv rj dvaXoyia, 81' 1'aou 8e xo npSxov xou 
xpixou y.siZ,ov fj, xal xo xexapxov xou exxou (iel^ov eaxai, 
xav laov, iaov, xav eXaxxov, eXaxxov. 

A i 1 A i 1 

B' 1 E' 1 

ri 1 z> ' 

'Eaxw xpia jxeye'dr] xa A, B, T xal aXXa auxolc; Xou xo 
TiXfj'dot; xa A, E, Z, auv8uo Xajjpavojjeva xal ev xQ auxw 
Xoyco, eax« 8e xexapaYjjevr) auxfiv f\ dvaXoyia, w<; jxsv xo 
A npo? xo B, ouxdx xo E npo? xo Z, cbc Se xo B Kpot; xo T, 
ouxw<; xo A 7ip6<; xo E, 81' laou Ss xo A xou T [ieT^ov eaxw 
Xey", oxi xal xo A xou Z (ieli^ov eaxai, xav Taov, laov, xav 
eXaxxov, eXaxxov. 

Tkel yap ]j.e%6\) eaxi xo A xou T, dXXo 8e xi xo B, xo 
A dpa Tipoc; xo B jjel^ova Xoyov e^ei f)7iep xo T 7ip6<; xo B. 
dXX'' [jev xo A Ttpoc; xo B, ouxW(; xo E npoq xo Z, w<; 
8e xo T Ttpoc; xo B, avduaXiv ouxwc; xo E icpoc; xo A. xal 
xo E apa Tipoc; xo Z [lei^ova Xoyov e)(ei f]Tcep xo E Ttpoc; xo 
A. Ttpoc; o 8e xo auxo jxel^ova Xoyov exei, exelvo eXaaaov 
eaxiv eXaaaov apa eaxl xo Z xou A- [jel^ov apa eaxl xo A 
xou Z. 6[jo(cl>c; 8r] 8elc;o[jev, oxi xav Taov fj xo A xCS T, laov 
eaxai xal xo A xQ Z, xav eXaxxov, eXaxxov. 

'Eav apa fj xpia jxeye'dr] xal aXXa auxolc; laa xo TtXfj'doc;, 
auv8uo Xajjpavojjcva xal ev iu auxw Xoyw, fj 8e xexa- 
payjjevr] auxwv r\ dvaXoyia, Si'iaou 8e xo TtpQxov xou xpixou 
jiel^ov fj, xal xo xexapxov xou exxou jiel£ov eaxai, xav taov, 



nitudes) having a ratio to the same (magnitude), that 
having the greater ratio is greater [Prop. 5.10]. Thus, 
D (is) greater than F. Similarly, we can show that even if 
A is equal to C then D will also be equal to F, and even 
if (A is) less (than C then D will also be) less (than F). 

Thus, if there are three magnitudes, and others of 
equal number to them, (being) also in the same ratio 
taken two by two, and (if), via equality, the first is greater 
than the third, then the fourth will also be greater than 
the sixth. And if (the first is) equal (to the third then the 
fourth will also be) equal (to the sixth) . And (if the first 
is) less (than the third then the fourth will also be) less 
(than the sixth) . (Which is) the very thing it was required 
to show. 

:: e : C then a = 7 as 5 = (. 

Proposition 21 1 

If there are three magnitudes, and others of equal 
number to them, (being) also in the same ratio taken two 
by two, and (if) their proportion (is) perturbed, and (if), 
via equality, the first is greater than the third then the 
fourth will also be greater than the sixth. And if (the first 
is) equal (to the third then the fourth will also be) equal 
(to the sixth) . And if (the first is) less (than the third then 
the fourth will also be) less (than the sixth). 

A i 1 Di 1 

B i 1 E i 1 

Ci 1 F i 1 

Let A, B, and C be three magnitudes, and D, E, F 
other (magnitudes) of equal number to them, (being) in 
the same ratio taken two by two. And let their proportion 
be perturbed, (so that) as A (is) to B, so E (is) to F, and 
as B (is) to C, so D (is) to E. And let A be greater than 
C, via equality. I say that D will also be greater than F. 
And if (A is) equal (to C then D will also be) equal (to 
F). And if (A is) less (than C then D will also be) less 
(than F). 

For since A is greater than C, and B some other (mag- 
nitude), A thus has a greater ratio to B than C (has) to 
B [Prop. 5.8]. But as A (is) to B, so E (is) to F. And, 
inversely, as C (is) to B, so E (is) to D [Prop. 5.7 corr.]. 
Thus, E also has a greater ratio to F than E (has) to D 
[Prop. 5.13]. And that (magnitude) to which the same 
(magnitude) has a greater ratio is (the) lesser (magni- 
tude) [Prop. 5.10]. Thus, F is less than D. Thus, D is 
greater than F. Similarly, we can show that even if A is 
equal to C then D will also be equal to F, and even if (A 
is) less (than C then D will also be) less (than F) . 



150 



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ELEMENTS BOOK 5 



laov, xdv eXaxxov, eXaxxov onsp eSei SeTi;ai. Thus, if there are three magnitudes, and others of 

equal number to them, (being) also in the same ratio 
taken two by two, and (if) their proportion (is) per- 
turbed, and (if), via equality, the first is greater than the 
third then the fourth will also be greater than the sixth. 
And if (the first is) equal (to the third then the fourth 
will also be) equal (to the sixth). And if (the first is) less 
(than the third then the fourth will also be) less (than the 
sixth) . (Which is) the very thing it was required to show. 

t In modern notation, this proposition reads that if a : j3 :: e : C and (3 : 7 :: <5 : e then a = 7 as <5 = (. 



x(3'. 

'Edv f) oTtoaaouv jieye'dr] xod aXXa auxolc; !'aa xo TtXfydoc;, 
auvSuo Xajjpavojjeva xod ev x£> auxw Xoycp, xod 81' taou ev 
xtp auxw Xoyco eaxai. 

a ' b ' r>— ' 

A 1 E ' Z' ' 

Hi 1 1 K' ' 1 1 Mi 1 1 

@l , , A 1 1 ' N' ' ' 

"Eaxw onoaaoOv ^sys'dr] xd A, B, T xod aXXa auxolc; I'aa 

xo TtXfj'doc; xa A, E, Z, auvSuo Xa^3otv6^eva ev xG auxcp 

Xoyw, oiQ \xev xo A npoc; xo B, ouxwc; xo A npbq xo E, cbc; 

8e xo B Ttpoc; xo T, ouxwc; xo E Ttpoc; xo Z- Xeyw, oxi xod 

81° I'aou ev xw auxw Xoyw eaxai. 

ElXf](p'f)w yap xwv jxev A, A ig&xu; TtoXXaicXdaia xa H, 

G, xwv Se B, E aXXa, a exu)(ev, ladxic; icoXXaiiXdaia xa K, 

A, xod exi xfiv T, Z aXXa, d exu^ev, ladxi? TioXXauXdaia xa 

M, N. 

Kal etxel eoxiv d>c; xo A Kpo<; xo B, ouxwg xo A 7ip6<; xo 
E, xal sl'XrjTCxai xov [lev A, A ladxi? KoXXanXdoia xa H, Q, 
xwv 8e B, E aXXa, a exu)(ev, ladxic KoXXanXdaia xa K, A, 
eaxiv dpa d>c xo H TT.poc xo K, ouxw<; xo O npbc, xo A. 8la 
xd auxd 8r] xal cbc; xo K Kpoc xo M, ouxw<; xo A 7ip6<; xo 
N. etc! ouv xpia ^teye'dr] eaxl xd H, K, M, xal aXXa auxolc 
laa xo nkfftoz xa 0, A, N, auv8uo Xa^pavo^teva xal ev tu 
auxai X6y«, 81' I'aou dpa, ei uuspexet to H xou M, uneps/si 
xal xo 6 xou N, xa! ei laov, Xaov, xal si eXaxxov, sXaxxov. 
xal eaxi xd ^.sv H, <d xGv A, A ladxic; noXXanXdoia, xd 8e 
M, N x£3v r, Z aXXa, a exvyev, ladxic noXXauXdaia. eaxiv 
dpa &>c, xo A npbc, xo T, oux«<; xo A Tipoc; xo Z. 

'Eav dpa f) oiraaaoOv [ieye'dr) xal aXXa auxolc; I'aa xo 
TiXfj-doc;, auvSuo Xajj.f3av6jj.sva ev ifi auxai Xoyco, xal 8l'Ioou 
ev x« auxG Xoyco eaxar onep e8ei 8eTc;ai. 



Proposition 22 f 

If there are any number of magnitudes whatsoever, 
and (some) other (magnitudes) of equal number to them, 
(which are) also in the same ratio taken two by two, then 
they will also be in the same ratio via equality. 

A 1 B' 1 C ' 

Di 1 Ei 1 F 1 

Gi 1 1 K' 1 1 ' Mi 1 ' 

Hi 1 1 Li 1 1 1 N 1 1 

Let there be any number of magnitudes whatsoever, 
A, B, C, and (some) other (magnitudes), D, E, F, of 
equal number to them, (which are) in the same ratio 
taken two by two, (so that) as A (is) to B, so D (is) to 
E, and as B (is) to C, so E (is) to F. I say that they will 
also be in the same ratio via equality. (That is, as A is to 
C, so D is to F.) 

For let the equal multiples G and H have been taken 
of A and D (respectively), and the other random equal 
multiples K and L of B and E (respectively), and the 
yet other random equal multiples M and N of C and F 
(respectively) . 

And since as A is to B, so D (is) to E, and the equal 
multiples G and H have been taken of A and D (respec- 
tively), and the other random equal multiples K and L 
of B and E (respectively), thus as G is to K, so H (is) to 
L [Prop. 5.4]. And, so, for the same (reasons), as K (is) 
to M, so L (is) to N. Therefore, since G, K, and M are 
three magnitudes, and H, L, and other (magnitudes) 
of equal number to them, (which are) also in the same 
ratio taken two by two, thus, via equality, if G exceeds M 
then H also exceeds N, and if (G is) equal (to M then H 
is also) equal (to N), and if (G is) less (than M then H is 
also) less (than N) [Prop. 5.20]. And G and H are equal 
multiples of A and D (respectively), and M and N other 
random equal multiples of G and F (respectively) . Thus, 
as A is to G, so D (is) to F [Def. 5.5]. 

Thus, if there are any number of magnitudes what- 
soever, and (some) other (magnitudes) of equal number 
to them, (which are) also in the same ratio taken two by 



151 



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ELEMENTS BOOK 5 



t In modern notation, this proposition reads that if a : /3 :: e : C and /3 : 

xy'. 

'Edv fj xpia ^eyei)/] xal aXXa auxolc; laa to TtXfj'doc; 
auvSuo Xa^pavo^ieva ev ifi auxco Xoycp, fj 8e xexapayjievr] 
auxcov fj dvaXoyia, xal 81' laou ev iS auxco Xoycp eaxai. 

A i 1 b 1 — 1 r ' ' 

A' — ' Ei 1 Z — ' 

f-Jl 1 1 1 @ I 1 1 1 J\_l 1 1 

K' — i — i — i Mi ' 1 N — i — 1 

'Eaxco xp(a ^.eyei!)/] xd A, B, T xal aXXa auxolc; laa xo 

TtX/j-doc; ouvSuo Xa[ipav6|jieva ev xcp auxcp Xoycp xa A, E, Z, 

eaxco Be xexapay^ievr] auxcov f) dvaXoyia, (be; ^iev xo A Ttpoc; 

xo B, ouxcoc; xo E Ttpoc; xo Z, coc; 8s xo B Ttpoc; xo T, ouxcoc; 

xo A Ttpoc; xo E- Xeyco, oxi eaxlv tlx; xo A Ttpoc; xo T, ouxcoc; 

xo A Ttpoc; xo Z. 

EiXfjcpdco xcov [itv A, B, A laaxic; TtoXXaTtXdoia xd H, O, 

K, xcov Be r, E, Z aXXa, a exu)(ev, lodxic; TtoXXaTtXdoia xd 

A, M, N. 

Kal ETtel lodxic; eaxl TtoXXaTtXdaia xd H, O xcov A, B, xd 
Be \iipr\ xolc; coaauxcoc; TtoXXaTtXaoioic; xov auxov zyz\ Xoyov, 
eaxiv apa cbc; xo A Ttpoc; xo B, ouxcoc; xo H Ttpoc; xo 6. 8id 
xa auxd 8f) xal coc; xo E Ttpoc; xo Z, ouxcoc; xo M Ttpoc; xo N - 
xa[ eaxiv cbc; xo A npoc; xo B, ouxcoc; xo E Ttpoc; xo Z- xal (be; 
apa xo H Ttpoc; xo 6, ouxcoc; xo M Ttpoc; xo N. xal STtei eaxiv 
cbc; xo B Ttpoc; xo V, ouxcoc; xo A Ttpoc; xo E, xal evaXXdJ; 
cbc; xo B Ttpoc; xo A, ouxcoc; xo T Ttpoc; xo E. xal ETtel xd 
Q, K xebv B, A laaxic; saxl TtoXXaTtXdaia, xd Bs [ispt] xolc; 
lodxic; TtoXXaTtXaaioic; xov auxov eyei Xoyov, eoxiv apa cbc; 
xo B Ttpoc; xo A, ouxcoc; xo Ttpoc; xo K. dXX' (be; xo B Ttpoc; 
xo A, ouxcoc; xo T Ttpoc; xo E- xal cbc; apa xo 6 Ttpoc; xo K, 
ouxcoc; xo r Ttpoc; xo E. TtdXiv, end xd A, M xcov T, E laaxic; 
eaxi TtoXXaTtXdaia, eoxiv apa cog xo T Ttpoc; xo E, ouxcoc; xo 
A Ttpoc; xo M. dXX' (be; xo T Ttpoc; xo E, ouxcoc; xo 9 Ttpoc; 
xo K' xal cbc; apa xo <d Ttpoc; xo K, ouxcoc; xo A Ttpoc; xo M, 
xal svaXXdc; cbc; xo <d Ttpoc; xo A, xo K Ttpoc; xo M. eBeix^T) 
6e xal cbc; xo H Ttpoc; xo 6, ouxcoc; xo M Ttpoc; xo N. ETtel 
ouv xpia jieye'dr) taxi xd H, <d, A, xal aXXa auxoic; laa xo 
TtXfj'dot; xd K, M, N auvBuo Xa[ipav6u.£va ev xco auxco Xoycp, 
xa[ eoxiv auxcov xexapayu.evr) f) dvaXoyia, 5i' I'oou apa, el 
UTtepexei xo H xou A, UTtepexei xal xo K xou N, xal ei laov, 
I'oov, xal ei eXaxxov, eXaxxov. xa[ eaxi xd [lev H, K xcov A, 
A lodxic; TtoXXaTtXdaia, xd 8e A, N xcov T, Z. eaxiv apa cbc; 
xo A Ttpoc; xo r, ouxcoc; xo A Ttpoc; xo Z. 

'Eav apa fj xpia ^icyc'dr) xal aXXa auxolc; loa xo TtXTjiSoc; 
auvBuo Xa|ipav6|jieva ev xco auxco Xoycp, fj 8e xexapay^ievr) 



two, then they will also be in the same ratio via equality. 
(Which is) the very thing it was required to show. 

:: C ■ V an d -y ■ S :: rj : 6 then a : 8 :: e : 9. 

Proposition 23 f 

If there are three magnitudes, and others of equal 
number to them, (being) in the same ratio taken two by 
two, and (if) their proportion is perturbed, then they will 
also be in the same ratio via equality. 

A 1 C 1 

Di — i E' ' F — i 

G 1 1 1 1 Hi — i — i — 1 Li 1 1 

Ki — i — i — i Mi 1 1 Ni — i 1 

Let A, B, and C be three magnitudes, and D, E and F 
other (magnitudes) of equal number to them, (being) in 
the same ratio taken two by two. And let their proportion 
be perturbed, (so that) as A (is) to B, so E (is) to F, and 
as B (is) to C, so D (is) to E. I say that as A is to C, so 
D (is) to F. 

Let the equal multiples G, H, and K have been taken 
of A, B, and D (respectively), and the other random 
equal multiples L, M, and TV of C, E, and F (respec- 
tively) . 

And since G and H are equal multiples of A and B 
(respectively), and parts have the same ratio as similar 
multiples [Prop. 5.15], thus as A (is) to B, so G (is) to 
H. And, so, for the same (reasons), as E (is) to F, so M 
(is) to N. And as A is to B, so E (is) to F. And, thus, as 
G (is) to H, so M (is) to N [Prop. 5.11]. And since as B 
is to C, so D (is) to E, also, alternately, as B (is) to D, so 
C (is) to E [Prop. 5.16]. And since H and K are equal 
multiples of B and D (respectively), and parts have the 
same ratio as similar multiples [Prop. 5.15], thus as B is 
to D, so H (is) to K. But, as B (is) to D, so C (is) to 
E. And, thus, as H (is) to K, so C (is) to E [Prop. 5.11]. 
Again, since L and M are equal multiples of C and E (re- 
spectively), thus as C is to E, so L (is) to M [Prop. 5.15]. 
But, as C (is) to E, so H (is) to K. And, thus, as H (is) 
to K, so L (is) to M [Prop. 5.11]. Also, alternately, as H 
(is) to L, so K (is) to M [Prop. 5.16]. And it was also 
shown (that) as G (is) to H, so M (is) to N. Therefore, 
since G, H, and L are three magnitudes, and K, M, and 
N other (magnitudes) of equal number to them, (being) 
in the same ratio taken two by two, and their proportion 
is perturbed, thus, via equality, if G exceeds L then K 
also exceeds N, and if (G is) equal (to L then K is also) 
equal (to N), and if (G is) less (than L then K is also) 
less (than N) [Prop. 5.21]. And G and K are equal mul- 
tiples of A and D (respectively), and L and N of G and 



152 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



auxcov f] dvaXoyia, xal 8i'foou ev xcp aOxcp Xoycp eaxai- onep F (respectively). Thus, as ^4 (is) to C, so D (is) to F 

eSei 8el£ai. [Def. 5.5]. 

Thus, if there are three magnitudes, and others of 
equal number to them, (being) in the same ratio taken 
two by two, and (if) their proportion is perturbed, then 
they will also be in the same ratio via equality. (Which is) 
the very thing it was required to show. 

t In modern notation, this proposition reads that if a : /3 :: e : £ and (3 : 7 :: 8 : t then a : 7 :: <5 : ( . 



x8'. 

'Edv uptOTOv npbz Seuxepov xov auxov e)(r) Xoyov xod 
xpixov 7ip6<; xexapxov, exn 8s xal TtefiTixov npbz Seuxepov 
xov auxov Xoyov xal exxov 7ip6<; xexapxov, xal auvxeiDev 
TtpGxov xal TtefjiTCXov Tipoc; Seuxepov xov auxov ec;ei Xoyov 
xal xpixov xal exxov 7ip6<; xexapxov. 

B 

A' ' 'H 

r> 1 

E 

A 1 1 © 

Z' 1 

IlpGxov yap xo AB Txpoc; Seupepov xo T xov auxov exexw 
Xoyov xal xpixov xo AE Tipoc; xexapxov xo Z, exexco Se xal 
Tte^iTtxov xo BH Tipoc; Seuxepov xo T xov auxov Xoyov xal 
exxov xo E0 Tipoc; xexapxov xo Z- Xeyw, oxi xal auvxcdev 
TipaSxov xal Tiefjmxov xo AH Tipoc; Seuxepov xo T xov auxov 
e^ei Xoyov, xal xpixov xal exxov xo A0 Tipoc; xexapxov xo 
Z. 

Tkel yap eaxiv cdc, xo BH Tipoc; xo T, ouxcoc; xo E0 Tipoc; 
xo Z, dvaTiaXiv apa oX xo T Tipoc; xo BH, ouxgk xo Z Tipoc; 
xo E9. etc el ouv eaxiv oX xo AB Tipoc; xo T, ouxgjc; xo AE 
Tipoc; xo Z, oX Se xo T Tipoc; xo BH, ouxck xo Z Tipoc; xo 
E6, Si' 'laou dpa eaxiv oX xo AB Tipoc; xo BH, ouxgjc; xo AE 
Tipoc; xo E0. xal ctccI Sir]pr)fisva ^teyei!)/] dvdXoyov eaxiv, xal 
auvxcdevxa dvdXoyov eaxai- eaxiv apa oX xo AH Tipoc; xo 
HB, ouxck xo A9 npbz xo 9E. eaxi Se xal 6X xo BH Tipoc; 
xo T, ouxcx xo E6 Tipoc; xo Z- 81' laou apa eaxiv cX xo AH 
Tipoc; xo T, ouxgjc; xo A9 Tipoc; xo Z. 

'Edv dpa upwxov Tipoc; Seuxepov xov auxov e^T) Xoyov 
xal xpixov Tipoc; xexapxov, e^T] Se xal tic^ltixov Tipoc; Seuxepov 
xov auxov Xoyov xal exxov Tipoc; xexapxov, xal auvxeiDev 
TtpaSxov xal Tiefjmxov Tipoc Seuxepov xov auxov e^ei Xoyov 
xal xpixov xal exxov Tipoc; xexapxov oTiep eSei Selc;ai. 



Proposition 24 f 

If a first (magnitude) has to a second the same ratio 
that third (has) to a fourth, and a fifth (magnitude) also 
has to the second the same ratio that a sixth (has) to the 
fourth, then the first (magnitude) and the fifth, added 
together, will also have the same ratio to the second that 
the third (magnitude) and sixth (added together, have) 
to the fourth. 

B 

A 1 1 G 

Ci 1 

E 

Di 1 H 

F 1 1 

For let a first (magnitude) AB have the same ratio to 
a second C that a third DE (has) to a fourth F. And let 
a fifth (magnitude) BG also have the same ratio to the 
second C that a sixth EH (has) to the fourth F. I say 
that the first (magnitude) and the fifth, added together, 
AG, will also have the same ratio to the second C that the 
third (magnitude) and the sixth, (added together), DH, 
(has) to the fourth F. 

For since as BG is to C, so EH (is) to F, thus, in- 
versely, as C (is) to BG, so F (is) to EH [Prop. 5.7 corr.]. 
Therefore, since as AB is to C, so DE (is) to F, and as C 
(is) to BG, so F (is) to EH, thus, via equality, as AB is to 
BG, so DE (is) to EH [Prop. 5.22]. And since separated 
magnitudes are proportional then they will also be pro- 
portional (when) composed [Prop. 5.18]. Thus, as AG is 
to GB, so DH (is) to HE. And, also, as BG is to C, so 
EH (is) to F. Thus, via equality as AG is to C, so DH 
(is) toF [Prop. 5.22]. 

Thus, if a first (magnitude) has to a second the same 
ratio that a third (has) to a fourth, and a fifth (magni- 
tude) also has to the second the same ratio that a sixth 
(has) to the fourth, then the first (magnitude) and the 
fifth, added together, will also have the same ratio to the 
second that the third (magnitude) and the sixth (added 



153 



ETOIXEIfiN z. 



ELEMENTS BOOK 5 



together, have) to the fourth. (Which is) the very thing it 
was required to show. 



t In modern notation, this proposition reads that if a : (3 :: 7 : 8 and e : f3 :: ( : 5 then a + e : f3 :: 7 + £ : 5. 



XZ 



'Edv Tsaaapa ^ley^TI dvdXoyov rj, to ^eyiaTov [auxcov] 
xal to eXd/iaTov Suo tcov Xoituov ^ei^ovd eoxiv. 



Ah 
Eh 



H 

— 1 — 



Zh 



— I — 



h A 



T5ax« xeaaapa (j.eye'd'ir) dvdXoyov id AB, IA, E, Z, a>c 
to AB Ttpoc; to TA, outw<; to E itpoi; to Z, eoto Se ^isyioTov 
[Lev ocutwv to AB, eXd)(icn:ov 8e to Z- Xeyw, oti Ta AB, Z 
twv TA, E [isi^ovd eaTiv. 

Kslcrdco yap T 9 E laov to AH, ifi Se Z laov to TO. 

'End [ouv] eaTiv £>c to AB 7tp6<; to TA, outck to E 
Ttpoc; to Z, laov 8s to [iev E t£) AH, to 8e Z tG r9, eaTiv 
dpa foe. to AB npoc; to TA, outgjc; to AH 7ipo<; to T0. 
xal ensi eaTiv foe, oXov to AB 7ip6<; 6Xov to TA, outca; 
dcpaipcdev to AH npoc; dcpaipcdev to I?0, xal Xoitcov dpa 
to HB npbe, Xoikov to OA ecrcai foe, oXov to AB Ttpoc; oXov 
to TA. ^ta^ov 8e to AB tou TA- uelCov dpa xal to HB toO 
9A. xal enel laov eaTi to [iev AH tw E, to Se T0 t« Z, 
Ta dpa AH, Z laa IcttI toI<; T9, E. xal [enel] eav [dviaou; 
I'aa TipoaT£i!)fi, Ta oXa dviad eaTiv, eav dpa] twv HB, 0A 
dviawv ovtwv xal jieiC^ovoc; toO HB tw ^tev HB TtpoaTcdfj 
Ta AH, Z, tG Se 9A TipooTei9f) Ta TO, E, auvdyeTai Ta 
AB, Z jieiCova twv TA, E. 

'Eav dpa Teaaapa jieye'dr) dvdXoyov fj, to jieyiaTOv 
auTSSv xal to eXd)(iaTov 860 tc5v XoituSv ^eii^ovd eaTiv. oitep 
eSei 8eT^ai. 



Proposition 25t 

If four magnitudes are proportional then the (sum of 
the) largest and the smallest [of them] is greater than the 
(sum of the) remaining two (magnitudes) . 

G 

1 



Ah 



Eh 



o 



H 

—1— 



t 1 1 

Let AB, CD, E, and F be four proportional magni- 
tudes, (such that) as AB (is) to CD, so E (is) to F. And 
let AB be the greatest of them, and F the least. I say that 
AB and F is greater than CD and E. 

For let AG be made equal to E, and CH equal to F. 

[In fact,] since as AB is to CD, so i? (is) to F, and 
£ (is) equal to AG, and F to CH, thus as is to CD, 
so AG (is) to CH. And since the whole AB is to the 
whole CD as the (part) taken away AG (is) to the (part) 
taken away CH, thus the remainder GB will also be to 
the remainder HD as the whole AB (is) to the whole CD 
[Prop. 5.19]. And AB (is) greater than CD. Thus, GB 
(is) also greater than HD. And since AG is equal to E, 
and GH to F, thus AG and F is equal to G# and E. 
And [since] if [equal (magnitudes) are added to unequal 
(magnitudes) then the wholes are unequal, thus if] AG 
and F are added to GB, and CH and £ to HD—GB 
and iJZ? being unequal, and GB greater — it is inferred 
that AB and F (is) greater than CD and E. 

Thus, if four magnitudes are proportional then the 
(sum of the) largest and the smallest of them is greater 
than the (sum of the) remaining two (magnitudes). 
(Which is) the very thing it was required to show. 



t In modern notation, this proposition reads that if a : (3 :: 7 : S, and a is the greatest and <5 the least, then a + 5 > /3 + 7. 



154 



ELEMENTS BOOK 6 



Similar Figures 



155 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



"Opoi. 

a'. "Ojioia ax^axa £U , duypa|j.[id eaxiv, oaa xdc; xe 
ycoviac; laac; e/ei xaxa ^nav xal xdc; Ticpl xdc; Taac; ycoviac; 
TtXeupdc; dvdXoyov. 

P'. 'Axpov xal [isaov Xoyov eu-deia xsx^ifjadai Xeyexai, 
oxav rj to? f) oXr) Ttpoc; xo ^si^ov x^trjua, ouxtoc; xo ^isTCov 
Ttpoc; xo eXaxxov. 

y'. "Ytyoz eaxi ndvxoc; axr^axoc; r) and xrjc; xopucprjc; era 
xr)v pdaiv xd-dexoc; dyo[ievr). 

a . 

Td xpiycova xdi xd 7tapaXX/]X6ypa|i^a xd bub xo auxo 
ix\)oc, ovxa izpbc, dXX/jXd eoxiv tbc; di pdaeic;. 

E A Z 




© H B T A K A 

"Eot(x> xpiycova piev xd ABr, ArA, TtapaXXr)X6ypa[i(jia 
8s xd Er, rZ bub xo auxo uijjoc; xo AE Xeyco, oxi eoxlv coc; 
f] Br pdoig Ttpoc; xt)v TA pdaic;, ouxcoc; xo ABr xpiycovov 
Ttpoc; xo ArA xpiycovov, xal xo Er 7iapaXXr)X6ypa^ov npbc, 
xo rZ TtapaXXr]X6ypa^ov. 

'ExpepXr]a"d« yap i] BA ecp' exdxepa xd [ispt] era xd 0, A 
a/jjiaa, xal xricrdcoaav xrj ^iev Br pdaei Taai [6oai8r)noxo0v] 
ai BH, H6, xrj 8s TA pdaei laai oaaiBrjTioxouv ai AK, KA, 
xal eTTe^euyjdcoaav ai AH, A6, AK, AA. 

Kal ETZsi laai eialv ai TB, BH, H9 dXXr|Xaic;, laa eaxi xal 
xd AOH, AHB, ABr xpiycova dXXr|Xoic;. 6aa7tXaaicov apa 
eaxlv f) OF pdaig xrjc; Br pdaecoc;, xoaauxanXdoiov eaxi 
xal xo A0r xpiycovov xou ABr xpiycovou. 8id xd auxa 
8/) oaarcXaaicov eaxlv f] Ar pdaic; xrjc; TA pdaecoc;, xoaau- 
xanXdaiov eaxi xal xo AAr xpiycovov xou ArA xpiycovou- 
xal et Xar] eaxlv f] 6r pdaic; xrj IA pdaei, laov eaxi xal xo 
AOr xpiycovov xtp ArA xpiytovto, xal si U7iepe)(ei f] ®T pdaic; 
xrjc; TA pdaecoc;, UTtepe/ei xal xo A0r xpiycovov xou ArA 
xpiycovou, xal ei eXdaatov, eXaaaov. xeaadptov 8r) ovxtov 
^.eye'dcov 8uo [iev pdaecov xcov Br, TA, Suo Se xpiycovcov 
xcov ABr, ArA elXrjuxai iadxic; uoXXaTiXdaia xrjc; (iev Br 
pdaecoc; xal xou ABr xpiycovou fj xe 0r pdaic; xal xo A6r 
xpiycovov, xrjc; 8s TA pdaecoc; xal xou AAr xpiycovou dXXa, 



Definitions 

1. Similar rectilinear figures are those (which) have 
(their) angles separately equal and the (corresponding) 
sides about the equal angles proportional. 

2. A straight-line is said to have been cut in extreme 
and mean ratio when as the whole is to the greater seg- 
ment so the greater (segment is) to the lesser. 

3. The height of any figure is the (straight-line) drawn 
from the vertex perpendicular to the base. 

Proposition V 

Triangles and parallelograms which are of the same 
height are to one another as their bases. 

E A F 




H G B C D K L 

Let ABC and ACD be triangles, and EC and CF par- 
allelograms, of the same height AC. I say that as base BC 
is to base CD, so triangle ABC (is) to triangle ACD, and 
parallelogram EC to parallelogram CF. 

For let the (straight-line) BD have been produced in 
each direction to points H and L, and let [any number] 
(of straight-lines) BG and GH be made equal to base 
BC, and any number (of straight-lines) DK and KL 
equal to base CD. And let AG, AH, AK, and AL have 
been joined. 

And since CB, BG, and GH are equal to one another, 
triangles AHG, AGB, and ABC are also equal to one 
another [Prop. 1.38]. Thus, as many times as base HC 
is (divisible by) base BC, so many times is triangle AHC 
also (divisible) by triangle ABC. So, for the same (rea- 
sons), as many times as base LC is (divisible) by base 
CD, so many times is triangle ALC also (divisible) by 
triangle ACD. And if base HC is equal to base CL then 
triangle AHC is also equal to triangle ACL [Prop. 1.38]. 
And if base HC exceeds base CL then triangle AHC 
also exceeds triangle ACL} And if (HC is) less (than 
CL then AHC is also) less (than ACL). So, their being 
four magnitudes, two bases, BC and CD, and two trian- 



156 



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ELEMENTS BOOK 6 



a exuxev, ladxic TtoXXarcXaaia fj xe Ar pdcnc xal to AAr 
xplyovov xal 8e8eixxai, oxi, ei uitepexei T) Or pdaic; xfj? IA 
pdaeoK, UTtepexei xal xo A0I 1 xpiycovov xou AAr xpiycivou, 
xai ei lor], I'aov, xdi ei eXaaawv, eXaaaov eaxiv dpa &>z rj 
Br pdai? upoc; xrjv EA pdaiv, oux«<; xo ABr xplyovov 7tpo<; 
xo ArA xplywvov. 

Kdi ctieI xou [ibv ABr xpiywvou BrnXdaiov eaxi xo Er 
7tapaXX/]X6ypa^ov, xou 5e ArA xpiyovou BinXdaiov eaxi 
xo Zr 7tapaXX/]X6ypa^ov, xd 8e jiepr) xou; cbaauxco? 710X- 
XaTtXaaioic; xov auxov exei Xoyov, eaxiv dpa aX xo ABr 
xplyovov Kp6<; xo ArA xpiycovov, ouxoc xo Er TtapaX- 
XrjXoypa^ov Ttpoc; xo ZT TtapaXXrjXoypa^ov. enel ouv 
eSeix^T), ok (j.ev rj Br pdai<; n;p6<; x/]v EA, ouxgx xo ABr 
xplyovov upoc; xo ArA xplywvov, cbc; Se xo ABr xpiycovov 
7ip6<; xo ArA xpiywvov, ouxco? xo Er 7iapaXXr]X6ypa^ov 
Tipoc xo rZ TtapaXXrjXoypa^iov, xal &<; dpa f] Br pdau; 7ip6<; 
xr]v TA pdaiv, ouxwc; xo Er 7tapaXX/]X6ypa[i^ov Ttpoc; xo Zr 
TtapaXXrjXoypa^ov. 

Td dpa xpiycova xal xd TtapaXXr)X6ypa^a xd O716 xo 
auxo utjioc; ovxa npoc dXXrjXd eaxiv cbt; al pdaeic oTtep e8ei 
8eTc;ai. 



gles, ABC and ACD, equal multiples have been taken of 
base BC and triangle ABC — (namely), base HC and tri- 
angle AH C — and other random equal multiples of base 
CD and triangle ADC — (namely), base LC and triangle 
ALC. And it has been shown that if base HC exceeds 
base CL then triangle AHC also exceeds triangle ALC, 
and if {HC is) equal (to CL then AHC is also) equal (to 
ALC), and if {HC is) less (than CL then AHC is also) 
less (than ALC). Thus, as base BC is to base CD, so 
triangle ABC (is) to triangle ACD [Def. 5.5]. And since 
parallelogram EC is double triangle ABC, and parallelo- 
gram FC is double triangle ACD [Prop. 1.34], and parts 
have the same ratio as similar multiples [Prop. 5.15], thus 
as triangle ABC is to triangle ACD, so parallelogram EC 
(is) to parallelogram FC. In fact, since it was shown that 
as base BC (is) to CD, so triangle ABC (is) to triangle 
ACD, and as triangle ABC (is) to triangle ACD, so par- 
allelogram EC (is) to parallelogram CF, thus, also, as 
base BC (is) to base CD, so parallelogram EC (is) to 
parallelogram FC [Prop. 5.11]. 

Thus, triangles and parallelograms which are of the 
same height are to one another as their bases. (Which is) 
the very thing it was required to show. 



t As is easily demonstrated, this proposition holds even when the triangles, or parallelograms, do not share a common side, and/or are not 
right-angled. 

t This is a straight-forward generalization of Prop. 1.38. 



P'- 

'Edv xpiyovou Ttapd ^uav xwv TtXeupov dx^rj xu; eu'deTa, 
dvdXoyov xejiel xd<; xou xpiywvou TtXeupdc;- xal edv ai xou 
xpiyovou TiXeupal dvdXoyov xu/jiSwcnv, i\ iia xcic, xo^iac; era- 
^euyvu^ievr] eu-dela Ttapd xr)v Xoin/jv eaxai xou xpiycovou 
TiXeupdv. 



A 




Tpiyiovou yap xou ABr 7xapdXXr)Xo<; [iia xcov nXeupfiv 
xfj Br fix'dw f) AE- Xeyw, oxi eaxiv «<; f] BA Ttp6<; x/]v AA, 
ouxdK f) TE Ttp6<; xrjv EA. 



Proposition 2 

If some straight-line is drawn parallel to one of the 
sides of a triangle then it will cut the (other) sides of the 
triangle proportionally. And if (two of) the sides of a tri- 
angle are cut proportionally then the straight-line joining 
the cutting (points) will be parallel to the remaining side 
of the triangle. 



A 




For let DE have been drawn parallel to one of the 
sides BC of triangle ABC. I say that as BD is to DA, so 
CE (is) to EA. 



157 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



'ETte^eu/'dcoaav yap ai BE, IA. 

laov apa eaxl to BAE xpiycovov iS IAE xpiycbvy 
era yap xrjc auxfjc; pdae(bc; eaxi xfjc; AE xal ev xalc auxaTc 
TtapaXXfjXoic; xdic; AE, BE dXXo 8e ti to AAE xpiywvov. xd 
8s Taa Ttpoc; to auxo xov auxov e)(ei Xoyov eaxiv apa (be; xo 
BAE xpiywvov Tipoc; xo AAE [xpiyovov], ouxwc; xo IAE 
xpiywvov Ttpoc; xo AAE xpiyovov. aXX' (be; [lev xo BAE 
xpiycovov Ttpoc; xo AAE, ouxwc; fj BA Tipoc; xrjv AA- utio 
yap xo auxo utjioc; ovxa xrjv aTto xou E stcl xrjv AB xd-dexov 
dyo^ievrjv Tipoc; dXXrjXd eiaiv (be; ai pdaeic- 5ia xa auxa 8rj 
(be; xo TAE xpiyovov Tipoc; xo AAE, ouxcx; fj TE Ttpoc; xrjv 
EA- xal (be; apa rj BA Tipoc; xf]v AA, ouxog fj TE Tipoc; xrjv 
EA. 

AXXd 8rj ai xou ABr xpiytbvou TtXeupal ai AB, Ar 
dvdXoyov xexurjadtoaav, (be; fj BA Tipoc; xrjv AA, ouxcoc 
f] TE Ttpoc; xrjv EA, xal CTte£e6)(i!)Gj f) AE- Xeyto, oxi 
TiapdXXrjXoc; eaxiv fj AE xfj Br. 

Tcbv yap auxfiv xaxaaxeuaa-devxwv, enei eaxiv (be; fj 
BA Tipoc; xrjv AA, ouxgjc; fj TE Tipoc; xrjv EA, dXX' (be; 
(lev f] BA Ttpoc; xrjv AA, ouxwc; xo BAE xpiywvov Ttpoc 
xo AAE xpiycovov, (be; Be rj TE Tipoc; xrjv EA, ouxwc; xo 
TAE xpiywvov Ttpoc; xo AAE xpiywvov, xal (be; apa xo 
BAE xpiywvov Ttpoc; xo AAE xpiywvov, ouxwc; xo TAE 
xpiyovov Ttpoc; xo AAE xpiywvov. exdxepov apa xwv BAE, 
TAE xpiywvwv Ttpoc; xo AAE xov auxov e)(ei Xoyov. taov 
apa eaxl xo BAE xpiywvov iu TAE xpiywvy xai eiaiv excl 
xrjc; auxfjc; pdae«<; xfjc; AE. xa 8e laa xpiywva xal era xfjc; 
auxfjc pdaewc ovxa xal ev xdic auxdic TtapaXXfjXoic eaxiv. 
TtapdXXrjXoc apa eaxlv fj AE xrj Br. 

'Eav apa xpiywvou itapa (liav xwv TtXeupwv dx'dfj xic 
euif)eTa, dvdXoyov xe^tel xac xou xpiywvou TtXeupdc xal eav 
ai xou xpiywvou itXeupai dvdXoyov xjirj'dGaiv, fj era xac; 
xoiudc eraCeuyvu^ievrj euifteTa Ttapa xrjv XoiTtfjv eaxai xou 
xpiyovou TtXeupdv oTtep e8ei 8e1c;ai. 



Y • 

'Eav xpiywvou fj ywvia 8[)(a xurj'dfj, fj 8e xe[Uvouaa xrjv 
ywviav eO'dela xejivrj xal xrjv pdaiv, xa xfjc; pdaewc; x^irj^axa 
xov auxov ec;ei Xoyov xdic; XoiTtdic; xou xpiywvou TtXeupaTc;- 
xal eav xd xfjc; pdoewc; x^irjjiaxa xov auxov g/rj Xoyov xalc; 
Xomalc; xou xpiycovou TtXeupaTc;, fj dito xfjc; xopucpfjc; era xrjv 
xojurjv era^euyvu[uevrj eu'dela 8i/a xejiel xrjv xou xpiycovou 
ycoviav. 

"Eot(x> xpiywvov xo ABr, xal xexjurjo'dw fj UTto BAr 
ywvia 8i)(a UTto xrjc; AA eu-deiac;- Xey«, oxi eaxlv (be; fj BA 
Ttpoc; xrjv TA, ouxoc; fj BA Ttpoc; xrjv Ar. 

"H)cd« yap Bid xou T xrj AA TtapdXXrjXoc; rj TE, xal 
8ia)cdeTaa fj BA aujiTtiTtxexcd auxfj xaxa xo E. 



For let BE and CD have been joined. 

Thus, triangle BDE is equal to triangle CDE. For 
they are on the same base DE and between the same 
parallels DE and BC [Prop. 1.38]. And ADE is some 
other triangle. And equal (magnitudes) have the same ra- 
tio to the same (magnitude) [Prop. 5.7]. Thus, as triangle 
BDE is to [triangle] ADE, so triangle CDE (is) to trian- 
gle ADE. But, as triangle BDE (is) to triangle ADE, so 
(is) BD to DA. For, having the same height — (namely), 
the (straight-line) drawn from E perpendicular to AB — 
they are to one another as their bases [Prop. 6.1]. So, for 
the same (reasons), as triangle CDE (is) to ADE, so CE 
(is) to EA. And, thus, as BD (is) to DA, so CE (is) to 
EA [Prop. 5.11]. 

And so, let the sides AB and AC of triangle ABC 
have been cut proportionally (such that) as BD (is) to 
DA, so CE (is) to EA. And let DE have been joined. I 
say that DE is parallel to BC. 

For, by the same construction, since as BD is to DA, 
so CE (is) to EA, but as BD (is) to DA, so triangle BDE 
(is) to triangle ADE, and as CE (is) to EA, so triangle 
CDE (is) to triangle ADE [Prop. 6.1], thus, also, as tri- 
angle BDE (is) to triangle ADE, so triangle CDE (is) 
to triangle ADE [Prop. 5.11]. Thus, triangles BDE and 
CDE each have the same ratio to ADE. Thus, triangle 
BDE is equal to triangle CDE [Prop. 5.9]. And they are 
on the same base DE. And equal triangles, which are 
also on the same base, are also between the same paral- 
lels [Prop. 1.39]. Thus, DE is parallel to BC. 

Thus, if some straight-line is drawn parallel to one of 
the sides of a triangle, then it will cut the (other) sides 
of the triangle proportionally. And if (two of) the sides 
of a triangle are cut proportionally, then the straight-line 
joining the cutting (points) will be parallel to the remain- 
ing side of the triangle. (Which is) the very thing it was 
required to show. 

Proposition 3 

If an angle of a triangle is cut in half, and the straight- 
line cutting the angle also cuts the base, then the seg- 
ments of the base will have the same ratio as the remain- 
ing sides of the triangle. And if the segments of the base 
have the same ratio as the remaining sides of the trian- 
gle, then the straight-line joining the vertex to the cutting 
(point) will cut the angle of the triangle in half. 

Let ABC be a triangle. And let the angle BAC have 
been cut in half by the straight-line AD. I say that as BD 
is to CD, so BA (is) to AC. 

For let CE have been drawn through (point) C par- 
allel to DA. And, BA being drawn through, let it meet 



158 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



E 




B A r 

Kal excel sic; rcapaXXf]Xouc; xdc; AA, Er eudela everceaev 
f) Ar, f] apa utio ArE ytovia Tar) eaxl xfj Oreo TAA. dXX' 
f] utio TAA xfj Oreo BAA urcoxeixai Tar)- xal f] utio BAA 
apa xfj utio ArE eaxiv Tar). rcdXiv, creel eic, TiapaXXfjXouc; 
xdc; AA, Er £1)15610 eveiieaev f] BAE, f) exxoc; ywvia f] Oreo 
BAA Tar) eaxl xfj evxoc; xfj utco AEr. sSei/i}/) 8s xal f] utio 
ArE xfj utco BAA Tar)- xal f) utco ArE dpa ycovia xfj utco 
AEr eaxiv Tar)- &axe xal TcXeupd f| AE TiXeupa xfj Ar eaxiv 
Xor\. xal eicel xpiycovou xou BrE Tcapd ^iav xwv TcXeupwv 
xf)v Er fjxxai f) AA, dvdXoyov apa eaxiv (be; f] BA Tcpoc; xf)v 
Ar, ouxwc; f) BA icpoc; xf)v AE. Tar] Be f) AE xfj Ar- cbc; apa 
f] BA Ttpoc; xf)v Ar, oux«<; f] BA icpoc; xf)v Ar. 

AXXd 8fj eaxco 6c; f] BA Tcpoc; xf]v Ar, ouxwc; f] BA icpoc 
xf]v Ar, xal iTieCeux'dw f] AA- Xeyto, oxi Biya xex^rjxai f| 
utco BAr ycovia utco xfjc; AA eufteiac;. 

Tt5v yap auxwv xaxaaxeuaa'devxwv, CTcei eaxiv «<; f) BA 
Tcpoc; xf]v Ar, ouxwc; f) BA Tcpoc; xf]v Ar, dXXd xal wc; f] BA 
Tcpoc; xr)v Ar, ouxwc; eaxiv f] BA Tcpoc; xf]v AE- xpiycovou 
yap xou BrE Tcapd ulav xf]v Er fjxxai f] AA- xal dbc; apa f] 
BA Tcpoc; xf]v Ar, ouxwc; f] BA Tcpoc; xf)v AE. Tar) apa f] Ar 
xfj AE- 6axe xal ywvia f) utco AEr xfj utco ArE eaxiv Tar). 
dXX' f] ^xev utco AEr xfj exxoc; xfj utio BAA [eaxiv] Tar], f) 
8e utco ArE xfj evaXXdc; xfj utco TAA eaxiv Tar)- xal f) utio 
BAA apa xfj utco TAA eaxiv Tar), f) apa utco BAr yovia 
8i)(a xexurjxai utio xfjc; AA eu-Mac;. 

'Eav dpa xpiycivou f] ywvia 8i)(a x^trydfj, f) 5e xe^ivouaa 
xf)V ywviav eu'dela xe^ivr] xal xf]v pdaiv, xa xfjc; pdaewc; 
x^ifjuaxa xov auxov ei;ei Xoyov xalc; Xomau; xou xpiywvou 
TiXeupaTc;- xal edv xa xfjc; pdaeioc; xjifj^iaxa xov auxov ey^r) 
Xoyov xalc XoiTtau; xou xpiywvou nXeupdic;, f) dTco xfjc; xo- 
pucpfjc; era xf)v xo^tfjv eTci£euyvutJievr] eu-dela 5ixa xeuvei xf)v 
xou xpiycovou ywviav onep eSei 8eTc;ai. 



t The fact that the two straight-lines meet follows because the sum of 



(CE) at (point) El 



E 




B DC 

And since the straight-line AC falls across the parallel 
(straight-lines) AD and EC, angle ACE is thus equal to 
CAD [Prop. 1.29]. But, (angle) CAD is assumed (to 
be) equal to BAD. Thus, (angle) BAD is also equal to 
ACE. Again, since the straight-line BAE falls across the 
parallel (straight-lines) AD and EC, the external angle 
BAD is equal to the internal (angle) AEC [Prop. 1.29]. 
And (angle) ACE was also shown (to be) equal to BAD. 
Thus, angle ACE is also equal to AEC. And, hence, side 
AE is equal to side AC [Prop. 1.6]. And since AD has 
been drawn parallel to one of the sides EC of triangle 
BCE, thus, proportionally, as BD is to DC, so BA (is) 
to AE [Prop. 6.2]. And AE (is) equal to AC. Thus, as 
BD (is) to DC, so BA (is) to AC. 

And so, let BD be to DC, as BA (is) to AC. And let 
AD have been joined. I say that angle BAG has been cut 
in half by the straight-line AD. 

For, by the same construction, since as BD is to DC, 
so BA (is) to AC, then also as BD (is) to DC, so BA is 
to A_E. For AD has been drawn parallel to one (of the 
sides) EC of triangle BCE [Prop. 6.2]. Thus, also, as 
BA (is) to AC, so BA (is) to AE [Prop. 5.11]. Thus, AC 
(is) equal to AE [Prop. 5.9]. And, hence, angle AEC 
is equal to ACE [Prop. 1.5]. But, AEC [is] equal to the 
external (angle) BAD, and ACE is equal to the alternate 
(angle) CAD [Prop. 1.29]. Thus, (angle) BAD is also 
equal to CAD. Thus, angle BAG has been cut in half by 
the straight-line AD. 

Thus, if an angle of a triangle is cut in half, and the 
straight-line cutting the angle also cuts the base, then the 
segments of the base will have the same ratio as the re- 
maining sides of the triangle. And if the segments of the 
base have the same ratio as the remaining sides of the 
triangle, then the straight-line joining the vertex to the 
cutting (point) will cut the angle of the triangle in half. 
(Which is) the very thing it was required to show. 

E and CAE is less than two right-angles, as can easily be demonstrated. 



159 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



See Post. 5. 

5'. 

Ttov iaoycovicov xpiytovtov dvdXoyov eiaiv ai TtXeupai al 
Tiepi xdc; loac ytoviac; xai o^oXoyoi ai utio xdc; iaac; ycoviac; 
UTioxdvouaai. 



z 




b r e 

"Ecrao iaoya>via xpiywva xd ABr, ArE iarjv £)(ovxa xrjv 
^iev utio ABr ycoviav xrj utio ArE, x/)v 8e utio BAr xfj utio 
TAE xai exi xrjv utio ArB xfj utio TEA- Xeyto, oxi xiov ABr, 
ArE xpiytoviov dvdXoyov eiaiv ai TiXsupai ai Tiepi xdc iaac; 
ycoviac; xai 6[i6Xoyoi ai (mo xdc; iaac; ycoviac; UTCoxdvouaai. 

Ksia'dco yap in euiiteiac; f) Br xfj TE. xai enel ai utio 
ABr, ArB ycoviai 8uo 6pi5cov sXdxxovec; eiaiv, iar] 8e 
f] utio ArB xrj utio AEr, ai apa utio ABr, AEr 8uo 
opiScov eXdxxovec; sioiv ai BA, EA apa £xpaXX6|ievai au^i- 
Tteaouvxai. expepXf|a , dcoaav xai au^iTiiTixexcoaav xaxd xo Z. 

Kai stcei lot) eaxiv f] utio ArE ycovia xrj utio ABr, 
TiapdXXrjXoc; eaxiv r) BZ xfj TA. TidXiv, ETcei iar) Eaxiv f] utio 
ArB xfj utio AEr, TiapdXXrjXoc; eaxiv f] AT xrj ZE. TiapaX- 
Xr]X6ypa[i|jiov apa eaxi xo ZAIA- iar) apa f) uev ZA xfj Ar, 
f] 8s Ar xrj ZA. xai ercei xpiycovou xou ZBE napd (iiav xrjv 
ZE rjxxai f) Ar, eaxiv apa cbc; f] BA Tipoc; xr)v AZ, ouxcoc; i\ 
Br Tipoc; xr]v TE. iar) 8e f) AZ xfj IA- 6c; apa f) BA Tipoc; xr)v 
IA, ouxcoc; f) Br Tipoc; xr)v TE, xai evaXXdi; cbc; f) AB Tipoc 
xr)v Br, ouxcoc; fj Ar Tipoc; xrjv TE. TidXiv, enei TiapdXXrjXoc; 
Eaxiv f) TA xfj BZ, Eaxiv apa cbc; f) Br Tipoc; xrjv TE, ouxcoc; 
f) ZA Tipoc; xrjv AE. i'arj 8s f) ZA xfj Ar- cbc; apa fj Br Tipoc; 
xrjv TE, ouxcoc; fj Ar Tipoc; xrjv AE, xai svaXXac; cbc; f) Br 
Tipoc; xrjv TA, ouxcoc; fj TE Tipoc; xrjv EA. eTiei oOv e8ei)($r) 
cbc [i£v f) AB Tipoc; xrjv Br, ouxcoc; fj Ar Tipoc; xrjv TE, cbc; 
8s rj Br Tipoc; xrjv TA, ouxcoc; fj TE Tipoc; xrjv EA, 8i' iaou 
apa cog f] BA Tipoc; xrjv Ar, ouxcoc; f] TA Tipoc; xrjv AE. 

Tcov apa iaoycovicov xpiycovcov dvdXoyov eiaiv ai TtXeupai 
ai Tiepi xdc; iaac; ycoviac; xai o^toXoyoi ai utio xdc; iaac; ycoviac; 
UTioxsivouaai- oTiep e8ei SeTc;ai. 



Proposition 4 

In equiangular triangles the sides about the equal an- 
gles are proportional, and those (sides) subtending equal 
angles correspond. 

F 




B C E 



Let ABC and DCE be equiangular triangles, having 
angle ABC equal to DCE, and (angle) BAG to CDE, 
and, further, (angle) ACB to CED. I say that in trian- 
gles ABC and DCE the sides about the equal angles are 
proportional, and those (sides) subtending equal angles 
correspond. 

Let BC be placed straight-on to CE. And since 
angles ABC and ACB are less than two right-angles 
[Prop 1.17], and ACB (is) equal to DEC, thus ABC 
and DEC are less than two right-angles. Thus, BA and 
ED, being produced, will meet [C.N. 5]. Let them have 
been produced, and let them meet at (point) F. 

And since angle DCE is equal to ABC, BF is parallel 
to CD [Prop. 1.28]. Again, since (angle) ACB is equal to 
DEC, AC is parallel to EE [Prop. 1.28]. Thus, FACD 
is a parallelogram. Thus, FA is equal to DC, and AC to 
FD [Prop. 1.34]. And since AC has been drawn parallel 
to one (of the sides) FE of triangle FBE, thus as BA 
is to AF, so BC (is) to CE [Prop. 6.2]. And AF (is) 
equal to CD. Thus, as BA (is) to CD, so BC (is) to CE, 
and, alternately, as AB (is) to BC, so DC (is) to CE 
[Prop. 5.16]. Again, since CD is parallel to BF, thus as 
BC (is) to CE, so FD (is) to DE [Prop. 6.2]. And FD 
(is) equal to AC. Thus, as BC is to CE, so AC (is) to 
DE, and, alternately, as BC (is) to CA, so Ci? (is) to 
ED [Prop. 6.2]. Therefore, since it was shown that as 
AB (is) to BC, so DC (is) to CE, and as BC (is) to CA, 
so Ci? (is) to ED, thus, via equality, as BA (is) to AC, so 
CD (is) to [Prop. 5.22]. 

Thus, in equiangular triangles the sides about the 
equal angles are proportional, and those (sides) subtend- 



160 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



z. 

'Edv 80o xptyova xdc; TiXeupdc; dvdXoyov e)(r), iaoyiovia 
saxai xd xpiycova xal foac; e^si xdc; yojviac;, uc p' ac; at 
6(i6Xoyoi TiXeupal UKOxeivouaiv. 

A A 




"Eaxco 860 xpiyova xd ABr, AEZ xdc; TiXsupdc; dvdXoyov 
e^ovxa, (be; ^lev xrjv AB Tipoc; xf)v Br, ouxwc; xf)v AE Tipoc; 
xf)v EZ, (be Ss xrjv Br Tipoc; xf)v FA, ouxwc; xf)v EZ Tipoc; 
xf)v ZA, xal ixi (be; xrjv BA Tipoc; xf]v Ar, ouxck xf]v EA 
Tipoc; xrjv AZ. Xeyw, oxi fooyoviov eaxi xo ABr xpiywvov 
iw AEZ xpiycovw xal foac; e^ouai xdc; ywviac;, Ocp' ac; ai 
o^ioXoyoi TiXeupal UTioxeivouaiv, xrjv ^tev utio ABr xfj utio 
AEZ, xf)v 8e utio BrA xfj utio EZA xal exi xrjv utio BAr 
xfj utio EAZ. 

Suveaxdxw yap Tipoc; xfj EZ eu-Ma xal xou; Tipoc; auxfj 
ot){ieioiz xou; E, Z xfj [lev utio ABr ywvia I'ar) f) utio ZEH, 
xfj 8s utio ArB for] f) utio EZH- XoiTtf] apa rj Tipoc; xw A 
Xomfj xfj Tipoc; tS H iaxiv I'ar). 

"Iaoycbvtov apa saxl xo ABr xpiywvov xw EHZ [xpiycbv- 
w] . xwv apa ABr, EHZ xpiycbvcov dvdXoyov eiaiv at TiXeupal 
at Tispl xdc; foac; yoviac; xal o^toXoyoi at utio xdc; foac; ywvtac; 
UTioxstvouaai- eaxiv apa (be; f] AB Tipoc; xrjv Br, [ouxwc;] 
f] HE Tipoc; xf]v EZ. dXX'' 6<; f] AB Tipoc; xf]v Br, ouxwc; 
UTioxsixai f) AE Tipoc; xf)v EZ- cbc; apa f) AE Tipoc; xf]v EZ, 
ouxox f] HE Tipoc; xf]v EZ. exaxepa apa xwv AE, HE Tipoc; 
xrjv EZ xov auxov s^si Xoyov Tar] apa saxtv f] AE xfj HE. 
81a xd auxd 8f] xal f] AZ xfj HZ eaxiv \ar\. kizzi ouv I'ar) eaxlv 
f] AE xfj EH, xoivf] 8e f) EZ, 860 8f) at AE, EZ 8ual xdic; HE, 
EZ foai etaiv xal pdaic; f] AZ pdaei xfj ZH [eaxiv] for) - yovia 
apa f) utio AEZ ywvia xfj utio HEZ eaxiv for), xal xo AEZ 
xplywvov xcp HEZ xpiycbvo foov, xal at Xoiiiat yoviai xau; 
XoiTidic; ywvtaic foai, ucp' ac; at foai TiXeupal UTioxeivouaiv. 
for] apa laxl xal f) \iev utio AZE ya>via xfj utio HZE, f) 8e 
utio EAZ xfj utio EHZ. xal etc el f) [Lev utio ZEA xfj Otio 
HEZ eaxiv for), dXX' f) utio HEZ xfj utio ABr, xal f) Otio 



ing equal angles correspond. (Which is) the very thing it 
was required to show. 

Proposition 5 

If two triangles have proportional sides then the trian- 
gles will be equiangular, and will have the angles which 
corresponding sides subtend equal. 

A D 




Let ABC and DEF be two triangles having propor- 
tional sides, (so that) as AB (is) to BC, so DE (is) to 
EF, and as BC (is) to CA, so EF (is) to FD, and, fur- 
ther, as BA (is) to AC, so ED (is) to DF. I say that 
triangle ABC is equiangular to triangle DEF, and (that 
the triangles) will have the angles which corresponding 
sides subtend equal. (That is), (angle) ABC (equal) to 
DEF, BCA to EFD, and, further, BAC to EDF. 

For let (angle) FEG, equal to angle ABC, and (an- 
gle) EFG, equal to ACB, have been constructed on the 
straight-line EF at the points E and F on it (respectively) 
[Prop. 1.23]. Thus, the remaining (angle) at A is equal 
to the remaining (angle) at G [Prop. 1.32]. 

Thus, triangle ABC is equiangular to [triangle] EGF. 
Thus, for triangles ABC and EGF, the sides about the 
equal angles are proportional, and (those) sides subtend- 
ing equal angles correspond [Prop. 6.4]. Thus, as AB 
is to BC, [so] GE (is) to EF. But, as AB (is) to BC, 
so, it was assumed, (is) DE to EF. Thus, as DE (is) to 
EF, so GE (is) to EF [Prop. 5.11]. Thus, DE and GE 
each have the same ratio to EF. Thus, DE is equal to 
GE [Prop. 5.9]. So, for the same (reasons), DF is also 
equal to GF. Therefore, since DE is equal to EG, and 
EF (is) common, the two (sides) DE, EF are equal to 
the two (sides) GE, EF (respectively). And base DF 
[is] equal to base FG. Thus, angle DEF is equal to 
angle GEF [Prop. 1.8], and triangle DEF (is) equal 
to triangle GEF, and the remaining angles (are) equal 
to the remaining angles which the equal sides subtend 
[Prop. 1.4]. Thus, angle DFE is also equal to GFE, and 



161 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



ABr dpa ytovia xfj Otco AEZ eaxiv for). 81a xd auxd 8f] xal 
f] Otio ArB xfj Otio AZE eaxiv Tar), xal exi f) Tipog xo A xfj 
Tipoc; xw A- iaoycbviov apa eaxl xo ABr xpiycovov xfi AEZ 
xpiyci>vq>. 

'Edv apa Buo xpiywva xag TiXeupdg dvdXoyov eyj), 
iaoycovia eaxai xd xpiywva xal laag e^ei xdg ycjviag, ucp' 
a<; al ojioXoyoi TtXeupal OTioxeivooaiv OTiep eSei Sel^ai. 



(angle) EDF to £G.F. And since (angle) FED is equal 
to GEF, and (angle) GEF to ^BC", angle ABC is thus 
also equal to DEF. So, for the same (reasons), (angle) 
ACB is also equal to DFE, and, further, the (angle) at A 
to the (angle) at D. Thus, triangle ABC is equiangular 
to triangle DEF. 

Thus, if two triangles have proportional sides then the 
triangles will be equiangular, and will have the angles 
which corresponding sides subtend equal. (Which is) the 
very thing it was required to show. 



Proposition 6 



'Edv 860 xpiytova ^uav ycoviav [iia ycovia forjv ey^r), Tiepl 
8s xd<; foa<; ywviag xdc; TiXeupdg dvdXoyov, iaoycovia eaxai 
xd xpiytova xal laac, e$ei xd<; ytoviag, ucp' ac, ai ojioXoyoi 
TiXeupal UTioxeivouaiv. 

A A 




b r 

'Eaxto 860 xpiyova xd ABr, AEZ jiiav ycoviav xfjv utio 
BAT [Lia ycovia xfj Otio EAZ forjv eyovxa, Tiepl Be xdc Xaac, 
ycoviac xac; TiXeupdg dvdXoyov, cb<; xf)v BA Tipoc; xfjv Ar, 
ooxcoc; xf]v EA npbz xf]V AZ- Xeyco, 6x1 iaoycoviov eaxi xo 
ABr xpiycovov xcp AEZ xpiy wvo xal iar]v e£ei xf)v Otio ABr 
ycoviav xfj hub AEZ, xfjv Be Otio ArB xfj bub AZE. 

Suveaxdxco yap Tipoc; xfj AZ eu-deia xal xoI<; Tipoc; ai) xfj 
at][ieioiz xolc; A, Z oTioxepa ^tev xcov bub BAT, EAZ for] 
f] Otio ZAH, xfj Be utio ArB Xar\ f) Otio AZH- XoiTif) apa f) 
Tipoc; xco B ycovia XoiTifj xfj Tipoc; iu H I'ar) eaxiv. 

'Iaoycoviov apa eaxl xo ABr xpiycovov iu AHZ 
xpiycovcp. dvdXoyov apa eaxiv cbc; f] BA Tipoc; xfjv Ar, ouxcoc; 
f] HA Tcpoc; xfjv AZ. UTioxeixai Be xal cbc; f] BA Tipoc; xfjv Ar, 
ooxcoc; f] EA Tipoc; xfjv AZ- xal cbc; apa f] EA Tipoc; xfjv AZ, 
ooxcoc; f) HA Tipoc; xf]v AZ. for] apa f] EA xfj AH- xal xoivf] 
f] AZ- Buo Sf] ai EA, AZ 8ual xaTc; HA, AZ Xaac, eiaiv xal 
ycovia f) Otio EAZ ycovia xfj Otio HAZ [eaxiv] Xat]- pdaic; 
apa f] EZ pdaei xfj HZ eaxiv Xat], xal xo AEZ xpiycovov iS 
HAZ xpiycovcp I'aov eaxiv, xal ai Xoiical ycoviai xau; XoiTidic; 
ycoviaic; Xaac, eaovxai, ucp' ac; I'aag TiXeupal UTioxeivouaiv. for] 
apa eaxiv f) [lev utio AZH xfj Otio AZE, f] Be Otio AHZ 



If two triangles have one angle equal to one angle, 
and the sides about the equal angles proportional, then 
the triangles will be equiangular, and will have the angles 
which corresponding sides subtend equal. 





B C 

Let ABC and DEF be two triangles having one angle, 
BAC, equal to one angle, EDF (respectively), and the 
sides about the equal angles proportional, (so that) as BA 
(is) to AC, so ED (is) to DF. I say that triangle ABC is 
equiangular to triangle DEF, and will have angle ABC 
equal to DEF, and (angle) ACB to DFE. 

For let (angle) FDG, equal to each of BAC and 
EDF, and (angle) DFG, equal to ACB, have been con- 
structed on the straight-line AF at the points D and F on 
it (respectively) [Prop. 1.23]. Thus, the remaining angle 
at B is equal to the remaining angle at G [Prop. 1.32]. 

Thus, triangle ABC is equiangular to triangle DGF. 
Thus, proportionally, as BA (is) to AC, so GD (is) to 
DF [Prop. 6.4]. And it was also assumed that as BA 
is) to AC, so ED (is) to DF. And, thus, as ED (is) 
to DF, so GD (is) to DF [Prop. 5.11]. Thus, ED (is) 
equal to DG [Prop. 5.9]. And DF (is) common. So, the 
two (sides) ED, DF are equal to the two (sides) GD, 
DF (respectively). And angle EDF [is] equal to angle 
GDF. Thus, base EF is equal to base GF, and triangle 
DEF is equal to triangle GDF, and the remaining angles 



162 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



xfj utio AEZ. dXX' f] Otio AZH xfj utco ArB eoxw for) - xal 
f) utco ArB apa xfj utco AZE eaxiv lay). UTCoxsraa 8s xal 
#1 utco BAr xf) utco EAZ I'ar)' xal Xoitct) apa f] Tcp6<; iu B 
Xomfj xf) Tcpoc; xt5 E lar) eaxlv laoycoviov apa eaxl xo ABr 
xplywvov tu AEZ xpiytovtp. 

Eav apa 860 xptyova uiav yoviav [iia yovia larjv EXTl; 
Tcepl 8e xa<; laa? ytovlac; xa<; TcXeupac; dvdXoyov, iaoyiovia 
eaxai xa xpiywva xal laac, e^ei xa; ytav[a<;, ucp' ac; ai 
o^ioXoyoi nXeupal UTCoxeivouaiv oTcep eSei SeT^ai. 



Eav Buo xpiycova [da\ ytoviav (iia ywvia I'orjv evt), 
Tcepl 8e dXXac; yioviac; xdg TcXeupdc; dvdXoyov, xaiv 8e 
XoitcGv exaxepav aua f]xoi eXdaaova fj [j.r) eXdaaova opdrjc, 
iaoycbvia eaxai xa xpiycova xal laac, e<;ei xa<; y«via<;, Tcepl 
ac dvdXoyov eiaiv ai TtXeupai. 




'Eaxw 860 xpiycova xa ABr, AEZ ^uav ycoviav [Lia 
ycovla iar]v e)(ovxa xfjv utco BAr xf] utco EAZ, Tcepl 8s 
dXXac; ycov(a<; xa<; utco ABr, AEZ xac; TcXeupdc; dvdXoyov, 
cog x/)v AB Tcpoc; xr)v Br, ouxcog xf)v AE Tcpoc; xf)v EZ, 
xcov 8e Xoitccov xcov Tcpoc; xolc; T, Z Tcpoxepov exaxepav 
d(ia eXdaaova op-drje Xeyco, oxi iaoycoviov eaxi xo ABr 
xpiycovov xco AEZ xpiycovcp, xal I'ar) eaxai f) utio ABr ycovia 
xrj utco AEZ, xal XoiTcfj BrjXovoxi f] Tcpoc; xcp T XoiTcfj xfj Tcpoc; 
xco Z Tar). 

EE yap dviaoc; eaxiv f) utio ABr ycovia xfj utio AEZ, 
fiia auxtov (lei^cov eaxiv. eaxw ^.ei^cov rj utco ABr. xal au- 
veaxdxco Tcpog xfj AB euiMa xal xco Tcpoc; auxfj arj^eico xcp 
B xrj utco AEZ ycovia Tar) rj utio ABH. 

Kal STtel Xar\ eaxiv f) [iev A ycovia xfj A, f) Be utco ABH 
xfj utco AEZ, Xoitct) apa f) utco AHB XoiTcfj xfj utco AZE 
eaxiv lay), iaoycoviov apa eaxl xo ABH xpiycovov ifi AEZ 



will be equal to the remaining angles which the equal 
sides subtend [Prop. 1.4]. Thus, (angle) DFG is equal 
to DFE, and (angle) DGF to DBF. But, (angle) DFG 
is equal to ACB. Thus, (angle) ACB is also equal to 
DFE. And (angle) BAG was also assumed (to be) equal 
to EDF. Thus, the remaining (angle) at B is equal to the 
remaining (angle) at E [Prop. 1.32]. Thus, triangle ABC 
is equiangular to triangle DEF. 

Thus, if two triangles have one angle equal to one 
angle, and the sides about the equal angles proportional, 
then the triangles will be equiangular, and will have the 
angles which corresponding sides subtend equal. (Which 
is) the very thing it was required to show. 

Proposition 7 

If two triangles have one angle equal to one angle, 
and the sides about other angles proportional, and the 
remaining angles either both less than, or both not less 
than, right-angles, then the triangles will be equiangular, 
and will have the angles about which the sides are pro- 
portional equal. 



A 




Let ABC and DEF be two triangles having one an- 
gle, BAG, equal to one angle, EDF (respectively), and 
the sides about (some) other angles, ABC and DEF (re- 
spectively), proportional, (so that) as AB (is) to BC, so 
DE (is) to EF, and the remaining (angles) at C and F, 
first of all, both less than right-angles. I say that triangle 
ABC is equiangular to triangle DEF, and (that) angle 
ABC will be equal to DEF, and (that) the remaining 
(angle) at C (will be) manifestly equal to the remaining 
(angle) at F . 

For if angle ABC is not equal to (angle) DEF then 
one of them is greater. Let ABC be greater. And let (an- 
gle) ABG, equal to (angle) DEF, have been constructed 
on the straight-line AB at the point B on it [Prop. 1.23]. 

And since angle A is equal to (angle) D, and (angle) 
ABG to DEF, the remaining (angle) ACB is thus equal 



163 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



xpiyovw. eaxiv dpa 6? f] AB Ttpoc; xrjv BH, ouxcoc; rj AE 
7ip6<; xr]v EZ. &>q 8e f] AE Ttp6<; xf]v EZ, [ouxgk] unoxeixai f) 
AB Ttpoc x/]v Br- f] AB apa Ttpoc; exaxepav xwv Br, BH xov 
auxov exei Xoyov iar] apa f) Br xrj BH. waxe xai ycovia rj 
jcpo<; iS r ywvia xfj utco BHr eaxiv I'ar). eXdxxwv 8e op^rjc; 
Oicoxeixai r] Ttpoc; xc5 T- eXdxxov dpa eaxiv opi^c; xal utio 
BHr- waxe f] ecpe^rjc; auxfj yovia f) utco AHB [ie(£«v eaxiv 
op^rjc;. xal eSe()cdr] Iar] ouaa xfj Ttpoc; xw Z- xal f\ npbq x£> Z 
dpa [ie(Cwv eaxiv op'drjc;. UTioxeixai Be eXdaawv op'drjc;- OTtep 
eaxiv dxoKov. oux dpa dviaoc; eaxiv f) utco ABr ywvia xfj 
utco AEZ- iar) dpa. eaxi 8e xal f] Ttpoc; x£> A iar] xfj Ttpoc; tu 
A- xal Xoircf] apa r) Ttpoc; iw T XoiTtfj xfj Ttpoc; xw Z Tar] eaxiv. 
iaoywviov dpa eaxi xo ABr xpiywvov x£> AEZ xpiywvw. 

AXXa 5r) TtdXiv UTtoxeia'dw exaxepa xwv Ttpoc; xou; T, Z \xr\ 
eXdaaov op'drjc- Xeyco TtdXiv, oxi xal ouxoc eaxiv iaoyciviov 
xo ABr xpiywvov iS AEZ xpiywvw. 

Twv yap auxfiv xaxaaxeuaai!)evxcL)v 6[ioitL>c Be(c;o^ev, 
6x1 Tar] eaxiv f) Br xrj BH- waxe xal ywvia f] npbq, xw T xfj 
utco BHr Tar] eaxiv. oux eXdxxov 8e op'dfjc r\ Ttpoc x£> E 
oux eXdxxcov apa op'drjc ouSe f) Otco BHr. xpiywvou 8r] xou 
BHr ai 8uo yoviai 860 op'dwv oux eiaiv eXdxxovec OTtep 
eaxiv dBuvaxov. oux dpa TidXiv dviaoc eaxiv f) utio ABr 
ywvia xfj utco AEZ- Tar] dpa. eaxi 8e xal f] npbq iw A xfj 
Ttpoc xw A Tar)- XoiTtr) apa f] Ttpoc xw T Xomfj xfj Ttpoc x<3 Z 
Tar] eaxiv. laoywviov apa eaxi xo ABr xpiywvov xw AEZ 
xpiywvw. 

'Eav dpa 860 xpiyova jiiav ywviav (iia yovia iarjv e)(r), 
Ttepi Be dXXac ycoviac xac TtXeupdc dvdXoyov, xov Be XoiTtov 
exaxepav d^ia eXdxxova fj \±r\ eXdxxova op^fji;, iaoywvia 
eaxai xd xpiywva xallaac e<;ei iac, yoviac, nepl ac, dvdXoyov 
eiaiv ai uXeupai- oiiep e8ei 8el5ai- 



'Eav ev 6pi3oy«v(w xpiywvcp aKO -zr\q op-dfj^ ycoviac sni 
x/]v pdaiv xdif)exo<; dx^fj, xd npbq xrj xa-dexcp xpiywva o^oid 
eaxi tu xe oXw xai aXkfikoiq. 

'Eaxw xpiywvov 6pif)oywviov xo ABr 6p{>f]v e^ov x/]v 
UTio BAr ywviav, xal f]xi3w duo xou A era xr]v Br xd-dexoc; 
f] AA- Xeyw, oxi b[Loiov eaxiv exdxepov xwv ABA, AAr 



to the remaining (angle) DFE [Prop. 1.32]. Thus, trian- 
gle ABG is equiangular to triangle DEF. Thus, as AB is 
to BG, so DE (is) to EF [Prop. 6.4]. And as L>£ (is) to 
EF, [so] it was assumed (is) AB to BC. Thus, has 
the same ratio to each of BC and BG [Prop. 5.11]. Thus, 
BC (is) equal to BG [Prop. 5.9]. And, hence, the angle 
at C is equal to angle BGC [Prop. 1.5]. And the angle 
at C was assumed (to be) less than a right-angle. Thus, 
(angle) BGC is also less than a right-angle. Hence, the 
adjacent angle to it, AGB, is greater than a right-angle 
[Prop. 1.13]. And (AGB) was shown to be equal to the 
(angle) at F. Thus, the (angle) at F is also greater than a 
right-angle. But it was assumed (to be) less than a right- 
angle. The very thing is absurd. Thus, angle ABC is not 
unequal to (angle) DEF. Thus, (it is) equal. And the 
(angle) at A is also equal to the (angle) at D. And thus 
the remaining (angle) at C is equal to the remaining (an- 
gle) at F [Prop. 1.32]. Thus, triangle ABC is equiangular 
to triangle DEF. 

But, again, let each of the (angles) at C and F be 
assumed (to be) not less than a right-angle. I say, again, 
that triangle ABC is equiangular to triangle DEF in this 
case also. 

For, with the same construction, we can similarly 
show that BC is equal to BG. Hence, also, the angle 
at C is equal to (angle) BGC. And the (angle) at C (is) 
not less than a right-angle. Thus, BGC (is) not less than 
a right-angle either. So, in triangle BGC the (sum of) 
two angles is not less than two right-angles. The very 
thing is impossible [Prop. 1.17]. Thus, again, angle ABC 
is not unequal to DEF. Thus, (it is) equal. And the (an- 
gle) at A is also equal to the (angle) at D. Thus, the 
remaining (angle) at C is equal to the remaining (angle) 
at F [Prop. 1.32]. Thus, triangle ABC is equiangular to 
triangle DEF . 

Thus, if two triangles have one angle equal to one 
angle, and the sides about other angles proportional, and 
the remaining angles both less than, or both not less than, 
right-angles, then the triangles will be equiangular, and 
will have the angles about which the sides (are) propor- 
tional equal. (Which is) the very thing it was required to 
show. 

Proposition 8 

If, in a right-angled triangle, a (straight-line) is drawn 
from the right-angle perpendicular to the base then the 
triangles around the perpendicular are similar to the 
whole (triangle), and to one another. 

Let ABC be a right-angled triangle having the angle 
BAC a right-angle, and let AD have been drawn from 



164 



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ELEMENTS BOOK 6 



xpiyovov oXcp tw ABr xal exi dXXf|Xoi<;. 



A 




b a r 

Tkei yap ' or l ecttIv f] Otto BAr xfj Otto AAB- opi}/) 
yap exaxepa' xal xoivf] xcov 8uo xpiycjvwv xou xe ABr xal 
xou ABA f] Tipoc; xfi B, Xoittt] apa f] Otto ArB XoiTtfj xfj 
utto BAA eaxiv Tar)- laoycoviov apa eaxl xo ABr xpiyiovov 
xG ABA xpiywvcp. eaxiv apa tbc; f) Br UTtoxeivouaa xfjv 
op'drjv xou ABr xpiywvou Ttpoc; xrjv BA UTtoxeivouaav xr)V 
opdrjv xou ABA xpiywvou, ouxw<; auxf] f] AB UTtoxeivouaa 
xf)V 7ip6<; xw r ywviav xou ABr xpiywvou npoQ xf)V BA 
UTtoxeivouaav xf]v tarjv xf)v Otto BAA xou ABA xpiywvou, 
xal exi f] AT Ttpoc; xf)v AA UTtoxeivouaav xf]v Ttpoc; ifi B 
ywviav xoivrjv xQv Buo xpiywvwv. xo ABr dpa xpiywvov 
x£S ABA xpiywvw laoycoviov xe eaxi xal xd<; rcepl xac; Xaac, 
yoviac; TtXeupdc; dvdXoyov ex £l - o\io\.o\ a\xa [eaxl] xo ABr 
xpiyovov tu ABA xpiycovcp. o^toiwc; §f] 8ei^o[iev, oxi xal 
tu AAr xpiywvw 6[loi6m eaxi xo ABr xpiywvov exdxepov 
dpa iwv ABA, AAr [xpiywvwv] 6\ioi6v eaxiv oXw iS ABr. 

Aeya> 8rj, oxi xal aXXVjXou; eaxiv o^toia xd ABA, AAr 
xpiyova. 

'Euel yap opiDf] f] utio BAA op'dfj xfj Otto AAr eaxiv 
tar), dXXd ^ir)v xal r] utto BAA xfj Ttpoc xCS T eBeix^r] Tar), 
xal XoiTtf] apa f] Ttpoc; x£5 B XoiTtfj xfj utio AAr eaxiv Tar)- 
laoycoviov apa eaxl xo ABA xplywvov x« AAr xpiyovo. 
eaxiv dpa cbc; f) BA xou ABA xpiywvou UTtoxeivouaa xf)V 
Otto BAA Tip6<; xf)v AA xou AAr xpiycovou UTtoxeivouaav 
xf)V Ttpoc; iS r Tarjv xfj Otto BAA, ouxwc; auxf] f) AA xou 
ABA xpiywvou UTtoxeivouaa xf]v Ttpoc; xw B ytoviav Ttpoc; 
xf)V Ar UTtoxeivouaav xfjv Otto AAr xou AAr xpiycivou 
Tarjv xfj Ttpoc; xw B, xal exi f] BA Ttpoc; xf)V Ar UTtoxeivouaai 
xdc; op-ddc;- o^toiov apa eaxl xo ABA xpiywvov iu AAr 
xpiywvw. 

'Eav apa ev 6pi!)oywviw xpiyovo dTto xfj? opi^fj^ ywviac; 
era xf)V pdaiv xd^exoc; dx'dfj, xd Ttpoc; xfj xa$exo xpiycova 
opioid eaxi xw xe oXw xal dXXf|Xoic; [oTtep e5ei Bel^ai]- 



A, perpendicular to BC [Prop. 1.12]. I say that triangles 
ABD and ADC are each similar to the whole (triangle) 
ABC and, further, to one another. 



A 




B DC 

For since (angle) BAG is equal to ADB — for each 
(are) right-angles — and the (angle) at B (is) common 
to the two triangles ABC and ABD, the remaining (an- 
gle) ACB is thus equal to the remaining (angle) BAD 
[Prop. 1.32]. Thus, triangle ABC is equiangular to tri- 
angle ABD. Thus, as BC, subtending the right-angle in 
triangle ABC, is to BA, subtending the right-angle in tri- 
angle ABD, so the same AB, subtending the angle at C 
in triangle ABC, (is) to BD, subtending the equal (an- 
gle) BAD in triangle ABD, and, further, (so is) AC to 
AD, (both) subtending the angle at B common to the 
two triangles [Prop. 6.4]. Thus, triangle ABC is equian- 
gular to triangle ABD, and has the sides about the equal 
angles proportional. Thus, triangle ABC [is] similar to 
triangle ABD [Dei. 6.1]. So, similarly, we can show that 
triangle ABC is also similar to triangle ADC. Thus, [tri- 
angles] ABD and ADC are each similar to the whole 
(triangle) ABC. 

So I say that triangles ABD and ADC are also similar 
to one another. 

For since the right-angle BDA is equal to the right- 
angle ADC, and, indeed, (angle) BAD was also shown 
(to be) equal to the (angle) at C, thus the remaining (an- 
gle) at B is also equal to the remaining (angle) DAC 
[Prop. 1.32]. Thus, triangle ABD is equiangular to trian- 
gle ADC. Thus, as BD, subtending (angle) BAD in tri- 
angle ABD, is to DA, subtending the (angle) at C in tri- 
angle ADC, (which is) equal to (angle) BAD, so (is) the 
same AD, subtending the angle at B in triangle ABD, to 
DC, subtending (angle) DAC in triangle ADC, (which 
is) equal to the (angle) at B, and, further, (so is) BA to 
AC, (each) subtending right-angles [Prop. 6.4]. Thus, 
triangle ABD is similar to triangle ADC [Def. 6.1]. 

Thus, if, in a right-angled triangle, a (straight-line) 
is drawn from the right-angle perpendicular to the base 



165 



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ELEMENTS BOOK 6 



U6pio\j.a. 

'Ex 8r) xouxou cpavepov, oxi edv £v opTJoyMviw xpiytovw 
duo xfj<; op-df^ y«vdi<; etii xr)v pdaic; xd-fJexoc; d^rj, f] 
dx'fMaa x£Sv xfjc; pdaeox x^ir^dxtov \±scr} dvdXoyov saxiv 
ojiep sSei BsT^ai. 

t In other words, the perpendicular is the geometric mean of the pieces. 



then the triangles around the perpendicular are similar 
to the whole (triangle), and to one another. [(Which is) 
the very thing it was required to show.] 

Corollary 

So (it is) clear, from this, that if, in a right-angled tri- 
angle, a (straight-line) is drawn from the right-angle per- 
pendicular to the base then the (straight-line so) drawn 
is in mean proportion to the pieces of the base.t (Which 
is) the very thing it was required to show 



0'. 

Trie Scddarjt; eO'ddae; xo Ttpoaxax'dev uepo<; dcpeXeiv. 




A Z B 

TEaxco f] So-dsiaa eO-deia f] AB- BsT 8f) xfje AB xo Ttpo- 
axaxiJEV [izpoz dcpeXeiv. 

'EKixsxdx'dw 8r] xo xpixov. [xdi] SiVj'dx 61 Tt< ^ TO ° A 
sMeia f\ AT ywviav izepie^ovaoi (iexd xfje AB xuxouaav xod 
slXr]Cpi&6j xuxov arj^islov em xfje Ar xo A, xdi xeiaOcoaav 
xfj AA I'aai al AE, EI\ xdi stte^sux'v}" #j Br, xdi 8id xoO A 
TtapdXXrjXoc; auxfj f^x'&w T) AZ. 

'End ouv xpiycovou xou ABr napd [iiav xfiv TtXeupfiv 
xf]v Br rjxxoa rj ZA, dvdXoyov dpa eaxlv &>z f\ TA 7tp6<; xf]V 
AA, ouxcoc; f] BZ itpoe xrjv ZA. SiTiXfj 8e #) IA xfje AA- 
SiTtXrj dpa xdi f\ BZ xrjc; ZA- xpmXrj dpa f) BA xfje AZ. 

Tfjc; dpa Bo-deiaTr]? sO'deia? xfj? AB xo emxax'dev xpixov 
^tepot; dcp/]pr]xai xo AZ- oitep s8ei Ttoifjaai. 



Proposition 9 

To cut off a prescribed part from a given straight-line. 




A F B 

Let AB be the given straight-line. So it is required to 
cut off a prescribed part from AB. 

So let a third (part) have been prescribed. [And] let 
some straight-line AC have been drawn from (point) A, 
encompassing a random angle with AB. And let a ran- 
dom point D have been taken on AC. And let DE and 
EC be made equal to AD [Prop. 1.3]. And let BC have 
been joined. And let DF have been drawn through D 
parallel to it [Prop. 1.31]. 

Therefore, since FD has been drawn parallel to one 
of the sides, BC, of triangle ABC, then, proportionally, 
as CD is to DA, so BE (is) to FA [Prop. 6.2]. And CD 
(is) double DA. Thus, BF (is) also double FA. Thus, 
BA (is) triple AF. 

Thus, the prescribed third part, AF, has been cut off 
from the given straight-line, AB. (Which is) the very 
thing it was required to do. 



i'. Proposition 10 

Trjv 8o$£iaav eCcdeiav dx|ir)xov xrj 8oi!)e[ar| xexur^evn To cut a given uncut straight-line similarly to a given 
b[LoioK xe^iav. cut (straight-line). 



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ELEMENTS BOOK 6 



r 




A Z H B 

"Eaxw f] ^.ev So-daaa sCWteTcx ax[ir]xo<; f] AB, f] 8e xe- 
x(i/)[j.£vr) f) Ar xaxa xa A, E ar)U£la, xal xeicrdojaav &axe 
ywviav xuxouaav nepiexeiv, xal eto^sux'&m rj TB, xal Bid 
xfiv A, E xfj Br TiapdXXrjXoi fj/iJioaav ai AZ, EH, Sid 8e 
xou A xfj AB TtapdXXiqXoc; fj)(#6j f) A0K. 

napaXXr]X6ypa[j.|iov dpa eaxlv sxdxspov x«v ZO, OB- 
Tor) dpa f) [isv A6 xfj ZH, f] 8s 6K xfj HB. xal ercel xpiyiovou 
xou AKr napd jiiav iwv nXsupwv xrjv Kr euifeTa rjxxai f] 
9E, dvdXoyov dpa Eaxlv foe, f) TE upoc; xrjv EA, ouxgjc; f) 
K6 Tipbc, x/)v 6A. lor) 8e f) ^ev K6 xfj BH, f] 8e OA xfj 
HZ. eaxiv dpa cb<; f] TE npoc, xrjv EA, ouxw<; f) BH npoc, xrjv 
HZ. TidXiv, ETtd xpiywvou xou AHE Ttapd |i(av xt5v TtXsupwv 
x/]v HE rjxxai t\ ZA, dvdXoyov dpa eaxlv 6<; r] EA 7tp6<; xrjv 
AA, ouxw<; f] HZ npoz xrjv ZA. eSdx'dr) 8e xal (be r) EE 
npoc; xr]v EA, ouxo<; f) BH npoc, xrjv HZ- eaxiv dpa «<; jiev 
f) TE npoc; xrjv EA, ouxcx fj BH npoc; xrjv HZ, cb<; 8e rj EA 
npoc xrjv AA, ouxck fj HZ 7tp6<; xrjv ZA. 

'H dpa So'deTaa su^eTa dx^trjxoc; rj AB xfj So-Marj eu-Ma 
xexjurj^evyj xfj Ar b[io'ux>z xexurjxai- omp s8ei Ttoirjaar 



la'. 

Auo So-deiafiv eu'deifiv xpixrjv dvdXoyov npoasupeTv. 

'Eaxwaav ai 8oif)eIaai [8uo eu'dETai] ai BA, Ar xal 
xsicrdtoaav ywvlav Ttepiixouaai xuxouaav. Bel 8rj xGv BA, 
Ar xpixrjv dvdXoyov upoaeupav. expepXfjcrdwaav yap em 
xd A, E arpeia, xal xslcxdo xfj Ar larj fj BA, xal e^eZ^X^ 
f] Br, xal Bid xou A rcapdXXrjXoc; auxfj fix&oi f] AE. 

Tkel ouv xpiycovou xou AAE Ttapd piiav xfiv nXeupfiv 
xrjv AE rjxxai fj Br, dvdXoyov egxiv &<; rj AB Ttpoc; xrjv 
BA, ouxwe; f) Ar Ttpoc; xrjv TE. Tar) Be f] BA xfj Ar. eoxiv 
dpa cb<; fj AB npog xrjv Ar, oux«<; rj Ar Ttpoc; xrjv TE. 



C 




A F G B 



Let AB be the given uncut straight-line, and AC a 
(straight-line) cut at points D and E, and let (AC) be 
laid down so as to encompass a random angle (with AB) . 
And let CB have been joined. And let DF and EG have 
been drawn through (points) D and E (respectively), 
parallel to BC, and let DHK have been drawn through 
(point) D, parallel to AB [Prop. 1.31]. 

Thus, FH and HB are each parallelograms. Thus, 
DH (is) equal to FG, and iJif to GB [Prop. 1.34]. And 
since the straight-line HE has been drawn parallel to one 
of the sides, KC, of triangle DKC, thus, proportionally, 
as CE is to ED, so KH (is) to #L> [Prop. 6.2]. And 
ifi? (is) equal to BG, and #L> to GF. Thus, as C£ is 
to ED, so £?G (is) to GF. Again, since FD has been 
drawn parallel to one of the sides, GE, of triangle AGE, 
thus, proportionally, as ED is to DA, so GF (is) to FA 
[Prop. 6.2]. And it was also shown that as CE (is) to 
ED, so SG (is) to GF. Thus, as CE is to £L>, so BG (is) 
to GF, and as ED (is) to FM, so GF (is) to FA. 

Thus, the given uncut straight-line, AB, has been cut 
similarly to the given cut straight-line, AC. (Which is) 
the very thing it was required to do. 

Proposition 11 

To find a third (straight-line) proportional to two 
given straight-lines. 

Let BA and AC be the [two] given [straight-lines], 
and let them be laid down encompassing a random angle. 
So it is required to find a third (straight-line) proportional 
to BA and AC. For let {BA and AC) have been produced 
to points D and F (respectively), and let BD be made 
equal to AC [Prop. 1.3]. And let BC have been joined. 
And let DE have been drawn through (point) D parallel 
to it [Prop. 1.31]. 

Therefore, since BC has been drawn parallel to one 
of the sides FF of triangle ADE, proportionally, as AB is 
to BD, so AC (is) to GF [Prop. 6.2]. And BD (is) equal 



167 



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ELEMENTS BOOK 6 



A 




Auo apa Bo^Eiawv EuiDeiwv xwv AB, Ar xplxT] dvdAoyov 
auxaac; 7tpoaeupT]xai f] TE- onep eSei uoifjoai. 



LP'. 

Tpiwv Soifteiacov sMeicov xsxdpxrjv dvdXoyov Ttpo- 
aeupeTv. 

A' ' 

B ' 



r 




A z 

"Eaxwaav ai So'der'iaai xpstc; euiDeTai ad A, B, F- SeT 8t) 
xwv A, B, r xexpdxrjv dvdXoyov ixpooeupslv. 

'ExxeiaiSwoav 860 eu'dslai ai AE, AZ ycoviav Tispis/oug- 
ai [xuxouaav] xt]v Otto EAZ- xal xsio-fko xfj Alar] r) AH, 
xf) 8s B laT) f) HE, xal exi xfj F far) f) A©- xal emZ^X^ e ' lar ]^ 
tt)c, HO KapdXXT)Xo<; auxfj ryyj}^ 81a xou E f) EZ. 

'EtcsI odv xpiyiovou xou AEZ rcapa jjiav xtjv EZ T)xxai f] 
HO, eaxiv apa tlx; f) AH npbc, xf)v HE, ouxioc; f) AO Ttpoc; 
xt)v OZ. laT) 8e f) piev AH xt) A, f) 8e HE xt) B, f) 8s AO xt) 
T- eaxiv apa &>z r) A upog xt]v B, ouxgj<; f) T npbc, xtjv OZ. 

Tpicov apa Scdeiawv eu$eic5v x«v A, B, T xexdpxT) 
dvdXoyov iipoosupTjxai f] OZ- onep e8ei ixoifjaaL. 



to AC. Thus, as is to AC, so AC (is) to CE. 



A 



D 




E 



Thus, a third (straight-line), C_E, has been found 
(which is) proportional to the two given straight-lines, 
AB and AC. (Which is) the very thing it was required to 
do. 

Proposition 12 

To find a fourth (straight-line) proportional to three 
given straight-lines. 

A' 1 

B' 1 




D H F 



Let A, B, and C be the three given straight-lines. So 
it is required to find a fourth (straight-line) proportional 
to A, B, and C. 

Let the two straight-lines DE and DF be set out en- 
compassing the [random] angle EDF. And let DG be 
made equal to A, and GE to B, and, further, DH to C 
[Prop. 1.3]. And GH being joined, let EF have been 
drawn through (point) E parallel to it [Prop. 1.31]. 

Therefore, since GH has been drawn parallel to one 
of the sides EF of triangle DEF, thus as DG is to GE, 
so DH (is) to HF [Prop. 6.2]. And DG (is) equal to A, 
and GE to B, and DH to C. Thus, as A is to £?, so C (is) 



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ELEMENTS BOOK 6 



Auo Bo-dsiawv eutJeiGv ^isaiqv dvdXoyov TtpoasupsTv. 




a Br 

"Eaxwaav ad BoiDeToai 8uo eu'dsTai ai AB, Br- Bel 8r) xSv 
AB, Br jiearjv dvdXoyov TtpoaEupsTv. 

Keicrdtoaav etc' eu'fkiac;, xal ycypdcp'ScL) etu xrjc; Ar 
f]^uxuxXiov to AAr, xal t^x'&co duo xoO B ar^dou xfj Ar 
su-Ma Ttpoc; opMc; f] BA, xal ETce^etix'fltoaav ai AA, Ar. 

'Etcei ev fj^ixuxXicp ycovia laxiv 7] Otco AAr, op'dr] eaxiv. 
xal etcei sv dpiJoycoviw xpiytovw xG AAr dico xrjc; opiDfjc; 
yoviac; era xr)v pdaiv xdiSsxoc rjxxai f\ AB, f] AB apa iSv 
xfj? pdoewi; x^ir^dxcov x£>v AB, Br \iear\ dvdXoyov ioxiv. 

Auo apa Bo'deiawv eu'deifiv xwv AB, Br \±£or} dvdXoyov 
icpoaeuprjxai f] AB- OTtep e8ei noifjaai. 



t In other words, to find the geometric mean of two given straight-lines. 

i5'. 

Toiv i'omv xs xal I'ooywviwv itapaXXr]Xoypdjj.(icov dvxi- 
Tteitovdaaiv ai icXsupal ai Ttepi xac; laac, ytoviac;- xal £>v ioo- 
ycoviwv iiapaXXrjXoypd^wv avxiTiETtovdaaiv ai TtXeupai ai 
Ttspl xac; laac; yoviac;, laa saxlv exeTva. 

"Eaxa> Taa xs xal iaoywvia TcapaXX/]X6ypa[i^.a xa AB, 
Br XaaLC, zyovxaL xac; Ttpoc; iu B ywviac;, xal xeicrdwaav etc' 
eMeiac; ai AB, BE- etc' cO^eiac; apa eiai xal ai ZB, BH. 
Xeyo, oxi xQv AB, Br avxiTieicovdaoiv ai TtXsupal ai Ttepi 
xac; laac; ywviac;, xouxeaxiv, oxi Eaxlv «<; f] AB Ttpoc; xr]v 
BE, ouxoc; f) HB Ttpoc; xrjv BZ. 



to HF. 

Thus, a fourth (straight-line), HF, has been found 
(which is) proportional to the three given straight-lines, 
A, B, and C. (Which is) the very thing it was required to 
do. 

Proposition 13 

To find the (straight-line) in mean proportion to two 
given straight-lines.^ 




A B C 

Let AB and BC be the two given straight-lines. So it 
is required to find the (straight-line) in mean proportion 
to AB and BC. 

Let {AB and BC) be laid down straight-on (with re- 
spect to one another), and let the semi-circle ADC have 
been drawn on AC [Prop. 1.10]. And let BD have been 
drawn from (point) B, at right-angles to AC [Prop. 1.11]. 
And let AD and DC have been joined. 

And since ADC is an angle in a semi-circle, it is a 
right-angle [Prop. 3.31]. And since, in the right-angled 
triangle ADC, the (straight-line) DB has been drawn 
from the right-angle perpendicular to the base, DB is 
thus the mean proportional to the pieces of the base, AB 
and BC [Prop. 6.8 corr.]. 

Thus, DB has been found (which is) in mean propor- 
tion to the two given straight-lines, AB and BC. (Which 
is) the very thing it was required to do. 



Proposition 14 

In equal and equiangular parallelograms the sides 
about the equal angles are reciprocally proportional. 
And those equiangular parallelograms in which the sides 
about the equal angles are reciprocally proportional are 
equal. 

Let AB and BC be equal and equiangular parallelo- 
grams having the angles at B equal. And let DB and BE 
be laid down straight-on (with respect to one another). 
Thus, FB and BC are also straight-on (with respect to 
one another) [Prop. 1.14]. I say that the sides of AB and 



169 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



e r 




A A 

Su^TtSTtXrjpcoa'dto yap to ZE TtapaXXr)X6ypa^i|iov. stisI 
ouv laov sera to AB TtapaXX/]X6ypa^ov iw Br TtapaXX/]- 
Xoypd^cp, aXXo Ss ti to ZE, sotiv apa (be; to AB Ttpoc; to 
ZE, outwc; to Br Ttpoc; to ZE. dXX'' (be; [Lev to AB upoc; 
to ZE, outwc; f] AB Ttpoc; ttjv BE, (be; 8s to Br Ttpoc; to 
ZE, outw<; f] HB Ttpoc; t/]v BZ- xal (be; apa f] AB Ttpoc; t/]v 
BE, outoc; f\ HB Ttpoc; ttjv BZ. twv apa AB, Br TtapaX- 
Xf]koyp6i[i[Ux>\i dvTiTtSTtovdaaiv al TtXsupal al Ttspl Tac; 1'aac; 
yovlac;. 

AXXd 5f] screw (be. f] AB Ttpoc; ttjv BE, outcoc; f\ HB Ttpoc; 
t/]v BZ- Xsyw, oil laov feaxl to AB TtapaXXr)X6ypa^ov iu 
Br TtapaXXr)Xoypd^«. 

Tksl yap sotiv to? #] AB Ttpoc; ttjv BE, outwc; #] HB 
Ttpoc; tt)v BZ, dXX'' (be; ^isv rj AB Ttpoc; ttjv BE, outcoc; to 
AB TtapaXX/]X6ypa^ov Ttpoc; to ZE TtapaXXr)X6ypa^ov, (be; 
8s f] HB Ttpoc; ttjv BZ, outmc; to Br TtapaXX/]X6ypa^ov 
Ttpoc; to ZE TtapaXX/]X6ypa^ov, xal (be; apa to AB Ttpoc; 
to ZE, oOtwc; to Br Tipoc; to ZE- laov apa taxi to AB 
TtapaXX/]X6ypa^ov tco Br TtapaXX/]Xoypdu[jicp. 

TGv apa 'lacov ts xal laoyovlcov TtapaXXrjXoypd^wv 
avTiTCSTiovdacuv al TtXsupal al Ttspl Tac; 1'aac; ywvlac;- xal SSv 
laoyovlov TtapaXXrjXoypd^cov dvTiTtSTtovdaaiv al TtXsupal 
al icspl Tac; 'laac; ywvlac;, I'aa scrav sxslva- OTtsp sBsi 8sTc;ai. 



is'. 

Tcov iacov xal ^uav ^tia \ar\v sxovtcov ywvlav Tpiycbvcov 
dvTiTtSTtovdaaiv al TtXsupal al Ttspl Tac; laac; ycovlac;- xal £>v 
[ilav (iia I'oiqv s)(6vt(ov ycovlav Tpiytbvtov avTiTtSTtovdaaiv al 
TtXsupal al Ttspl Tac; 'laac; ycovlac;, laa sotIv sxslva. 

'Earco !aa Tplycova to ABr, AAE jilav [Lia iar)v sxovTa 
ycovlav ttjv uno BAr Tfj utio AAE- Xsyto, oti tGv ABr, 
AAE Tpiyovwv dvTiTtSTtovdaaiv al TtXsupal al Ttspl Tac; i'aac; 
ycovlac;, toutsctuv, oti sgtIv (be; i] FA Ttpoc; t/]v AA, outcoc; 



BC about the equal angles are reciprocally proportional, 
that is to say, that as DB is to BE, so GB (is) to BF. 

E C 




A D 

For let the parallelogram FE have been completed. 
Therefore, since parallelogram AB is equal to parallelo- 
gram BC, and FE (is) some other (parallelogram), thus 
as (parallelogram) AB is to FE, so (parallelogram) BC 
(is) to FE [Prop. 5.7]. But, as (parallelogram) AB (is) to 
FE, so DB (is) to BE, and as (parallelogram) BC (is) to 
BB, so GB (is) to [Prop. 6.1]. Thus, also, as DB (is) 
to BE, so GB (is) to BF. Thus, in parallelograms AB 
and BC the sides about the equal angles are reciprocally 
proportional. 

And so, let DB be to BE, as GB (is) to BF. I say that 
parallelogram AB is equal to parallelogram BC. 

For since as DB is to BE, so GB (is) to BF, but as 
BB (is) to BE, so parallelogram AB (is) to parallelo- 
gram FE, and as GB (is) to BF, so parallelogram BC 
(is) to parallelogram FE [Prop. 6.1], thus, also, as (par- 
allelogram) AB (is) to FE, so (parallelogram) BC (is) 
to FE [Prop. 5.11]. Thus, parallelogram AB is equal to 
parallelogram BC [Prop. 5.9]. 

Thus, in equal and equiangular parallelograms the 
sides about the equal angles are reciprocally propor- 
tional. And those equiangular parallelograms in which 
the sides about the equal angles are reciprocally propor- 
tional are equal. (Which is) the very thing it was required 
to show. 

Proposition 15 

In equal triangles also having one angle equal to one 
(angle) the sides about the equal angles are reciprocally 
proportional. And those triangles having one angle equal 
to one angle for which the sides about the equal angles 
(are) reciprocally proportional are equal. 

Let ABC and ADE be equal triangles having one an- 
gle equal to one (angle), (namely) BAC (equal) to DAE. 
I say that, in triangles ABC and ADE, the sides about the 



170 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



f] EA Ttpoc x/]v AB. 



b r 




A E 

Ksia'dw yap uoxe en su-deiac slvai xrjv EA xfj AA- eti' 
eu-Qeiac, apa eaxl xal f] EA xrj AB. xal ETteCeux'dw f] BA. 

Ekei ouv laov taxi xo ABr xpiycovov xw AAE xpiywvw, 
aXXo 8e xi xo BAA, saxiv apa cbc xo TAB xpiywvov Ttpoc 
xo BAA xpiycovov, ouxwc xo EAA xpiywvov Ttpoc xo BAA 
xpiyovov. dXX' cbc ^.ev xo TAB Ttpoc xo BAA, ouxwc f] 
TA Ttpoc x/]v AA, 6? Ss xo EAA Ttpoc xo BAA, ouxtoc f] 
EA Ttpoc x/]v AB. xal <i>c apa f] TA Ttpoc x/]v AA, ouxoc 
f] EA Ttpoc x/]v AB. xwv ABr, AAE apa xpiywvwv dvxi- 
TiETtovdaaiv ai TtXsupal al Ttspl xdc taac ycoviac. 

AXXa 8r) avxiKETiovdsxcoaav ai TtXeupal xwv ABr, AAE 
xpiycivov, xal saxco wc f] EA Ttpoc xrjv AA, ouxmc f] EA 
Tipoc xr]v AB- Xsyw, oxi taov eaxl xo ABr xpiywvov iw 
AAE xpiyova). 

'Emc'sux'deiaTjc yap TtdXiv xrjc BA, stcei eaxiv «c f] TA 
Ttpoc; xf]v AA, ouxwc r) EA Ttpoc xrjv AB, dXX'' cbc jiev 
f\ TA Tipoc x/]v AA, ouxcoc xo ABr xplywvov Ttpoc xo 
BAA xpiywvov, cbc 5e rj EA Ttpoc xrjv AB, ouxcoc xo EAA 
xpiyovov Ttpoc xo BAA xpiywvov, (be apa xo ABr xplywvov 
Ttpoc xo BAA xpiywvov, oux«c xo EAA xpiywvov Ttpoc 
xo BAA xpiywvov. exdxepov apa xwv ABr, EAA Ttpoc 
xo BAA xov auxov Xoyov. lacov apa eaxl xo ABr 
[xpiycovov] xo EAA xpiyovco. 

Twv apa lawv xal ^.[av \ua larjv £)(6vxcl>v ywvlav 
xpiycbvov dvxiTtSTtovdaaiv ai TtXeupal ai Ttepl xdc I'aac 
ycoviac xal 5c ^iav (iia larjv e)(6vx«v ywviav xpiyovov 
avxiTteitov^aaiv ai TtXeupal ai Ttepl xdc laac, yoviac, exelva 
taa eaxlv oTtep eSsi 5eTcai. 



'Edv xeaaapec euiDelai dvdXoyov Saiv, xo utio xfiv 
dxptov Tt£pi£)(6|jievov op-doytbviov i'aov Eaxl ifi utio xfiv 
(jifawv Tiepie/o^evo opiJoywviw- xav xo utio xGv axpwv 



equal angles are reciprocally proportional, that is to say, 
that as CA is to AD, so EA (is) to AB. 

B C 




D E 

For let CA be laid down so as to be straight-on (with 
respect) to AD. Thus, EA is also straight-on (with re- 
spect) to AB [Prop. 1.14]. And let BD have been joined. 

Therefore, since triangle ABC is equal to triangle 
ADE, and BAD (is) some other (triangle), thus as tri- 
angle CAB is to triangle BAD, so triangle EAD (is) to 
triangle BAD [Prop. 5.7]. But, as (triangle) CAB (is) 
to BAD, so CA (is) to AD, and as (triangle) EAD (is) 
to BAD, so EA (is) to AB [Prop. 6.1]. And thus, as CA 
(is) to AD, so EA (is) to Ai?. Thus, in triangles ABC and 
ADE the sides about the equal angles (are) reciprocally 
proportional. 

And so, let the sides of triangles ABC and ADE be 
reciprocally proportional, and (thus) let C A be to AD, 
as EA (is) to AB. I say that triangle ABC is equal to 
triangle ADE. 

For, BD again being joined, since as CA is to AD, so 
£A (is) to AB, but as CA (is) to AD, so triangle ABC 
(is) to triangle BAD, and as _EA (is) to AB, so triangle 
£A£> (is) to triangle BAD [Prop. 6.1], thus as triangle 
ABC (is) to triangle BAD, so triangle EAD (is) to tri- 
angle BAD. Thus, (triangles) ABC and EAD each have 
the same ratio to BAD. Thus, [triangle] ABC is equal to 
triangle EAD [Prop. 5.9]. 

Thus, in equal triangles also having one angle equal to 
one (angle) the sides about the equal angles (are) recip- 
rocally proportional. And those triangles having one an- 
gle equal to one angle for which the sides about the equal 
angles (are) reciprocally proportional are equal. (Which 
is) the very thing it was required to show. 

Proposition 16 

If four straight-lines are proportional then the rect- 
angle contained by the (two) outermost is equal to the 
rectangle contained by the middle (two) . And if the rect- 



171 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



Ti£pi£)(6^£vov op'doytoviov loov rj tw (mo xcov (isawv Tiepie- 
/o|ji£vw op-doytoviw, ai xeaaapec; eO'deTai dvaXoyov eaovxai. 



H 



A 



e 



b r 



angle contained by the (two) outermost is equal to the 
rectangle contained by the middle (two) then the four 
straight-lines will be proportional. 

H 



B C 



D 



E' ' 

"Ecraoaocv xeaaapec; eu'delai dvaXoyov ai AB, TA, E, Z, 
cbc; f) AB Tipoc; tt)v TA, ouxgjc; f) E Tipoc; xf)v Z' Xeyio, oxi 
xo utio xcov AB, Z 7iepi£)(6[i£vov op'doywviov laov eaxi tu 
uno xov TA, E Tiepiexo^tevw opTJoyMviw. 

"Hx^waav [yap] duo xwv A, T ar^eitov xdic; AB, TA 
eu-deiaic; Tipoc; op-ddc; ai AH, TO, xai xeiaiDw xfj (jiev Z tar) 
f] AH, xfj Se E i'ar] f] TO. xai auuTieTiXir]pt5a , d« xd BH, A6 
TiapaXX/]X6ypa^a. 

Kai CTiei eaxiv w<; f] AB Tipoc; xrjv TA, ouxwc; f] E Ttpoc 
xf)v Z, iar) 5e f] ^tev E xfj r0, f] 8e Z xfj AH, eaxiv dpa w<; 
f] AB Tipoc; xf)v TA, ouxck f] T0 Tipoc; xf)v AH. xfiv BH, 
A0 apa TiapaXXrjXoypdu^iMv dvxiTiCTiovdaaiv ai TiXeupal ai 
Tcspl xac; laac; yoviac;. Sv 8e iaoytoviwv TiapaXXr)Xoypd^(jiov 
dvxiTienovdaaiv ai TiXeupai ai Tiepi xac; i'aac; ytovdic;, laa eaxiv 
exelva- laov apa eaxi xo BH TiapaXX/]X6ypa^ov x<5 A6 
7iapaXX/]Xoypd^^.CL>. xai eaxi xo jiev BH xo uno xCSv AB, Z' 
iar] yap f] AH xfj Z- xo Se A0 xo utio xfiv TA, E- Xar] yap f) 
E xfj T9- xo apa utio twv AB, Z Tiepiexo^evov op'doytoviov 
laov eaxi x£5 uno xwv TA, E Txepiexo^evw op-doytoviw. 

AXXd 8f] xo utio xfiv AB, Z Tiepiexo^tevov op-doycoviov 
laov eaxw xw utio twv TA, E Tiepiexo^tevo 6pi9oycovicp. 
Xeyo, oxi ai xeaaapec eu'delai dvaXoyov eaovxai, dbc; f] AB 
Tipoc; xf)v TA, ouxoc; f) E Tipoc xf)V Z. 

Tcov yap auxfiv xaxaaxeuaa-devxcov, etxel xo utio twv 
AB, Z laov eaxi xG utio iuv TA, E, xai eaxi xo ^.ev utio 
x£Sv AB, Z xo BH- Iar) yap eaxiv f) AH xfj Z- xo 8e utio icov 
TA, E xo A9- i'ar) yap f) T8 xfj E- xo apa BH i'aov eaxi 
tu A0. xai eaxiv iaoywvia. iuv 8e lawv xai iaoyovicov 
TiapaXX/jXoypd^tov dvxiTienovdaaiv ai TiXeupal ai Tiepi xac; 
laac; ywviac;. eaxiv apa &>z f] AB Tipoc; xf]v TA, ouxoc f] T& 
Tipoc; xf)V AH. iar] 8e f) ^tev r0 xfj E, f) 8e AH xfj Z- eaxiv 
apa cbc; f] AB Tipoc; xf)V TA, ouxw<; f\ E Tipoc xfjv Z. 

'Edv apa xeaaapec euiJeTai dvaXoyov waiv, xo utio xwv 
axpwv nepiexo^ievov op-doycoviov i'aov eaxi xw utio tuv 
^ifawv nepiexo^ievw op-doyMviw- xav xo utio xwv axpwv 
Tiepiex6[ievov op-doywviov laov fj xQ utio xwv [icawv iiepie- 
Xo^tevw op'doywvico, ai xeaaapec eu^elai dvaXoyov eaovxar 
oTiep e8ei 8eT^ai. 



Ei 1 F' 1 

Let AB, CD, E, and F be four proportional straight- 
lines, (such that) as AB (is) to CD, so E (is) to F. I say 
that the rectangle contained by AB and F is equal to the 
rectangle contained by CD and E. 

[For] let AG and CH have been drawn from points 
A and C at right-angles to the straight-lines AB and CD 
(respectively) [Prop. 1.11]. And let AG be made equal to 
F, and CH to E [Prop. 1.3]. And let the parallelograms 
BG and DH have been completed. 

And since as AB is to CD, so E (is) to F, and E 
(is) equal CH, and F to AG, thus as AB is to CD, so 
C# (is) to AG. Thus, in the parallelograms BG and Di? 
the sides about the equal angles are reciprocally propor- 
tional. And those equiangular parallelograms in which 
the sides about the equal angles are reciprocally propor- 
tional are equal [Prop. 6.14]. Thus, parallelogram BG 
is equal to parallelogram DH. And BG is the (rectangle 
contained) by AB and F. For AG (is) equal to F. And 
DH (is) the (rectangle contained) by CD and E. For E 
(is) equal to CH. Thus, the rectangle contained by AB 
and F is equal to the rectangle contained by CD and E. 

And so, let the rectangle contained by AB and F be 
equal to the rectangle contained by CD and E. I say that 
the four straight-lines will be proportional, (so that) as 
AB (is) to CD, so E (is) to F. 

For, with the same construction, since the (rectangle 
contained) by AB and F is equal to the (rectangle con- 
tained) by CD and E. And BG is the (rectangle con- 
tained) by AB and F. For AG is equal to F. And DH 
(is) the (rectangle contained) by CD and E. For CiJ 
(is) equal to E. BG is thus equal to DH. And they 
are equiangular. And in equal and equiangular parallel- 
ograms the sides about the equal angles are reciprocally 
proportional [Prop. 6.14]. Thus, as AB is to CD, so CH 
(is) to AG. And CH (is) equal to E, and AG to F. Thus, 
as AB is to CD, so E (is) to F. 

Thus, if four straight-lines are proportional then the 
rectangle contained by the (two) outermost is equal to 
the rectangle contained by the middle (two) . And if the 
rectangle contained by the (two) outermost is equal to 



172 



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ELEMENTS BOOK 6 



'Edv xpeu; eO'delai dvdXoyov Gmv, to Oko xGv dxptov 
Tiepiexo^ievov op-doyGviov laov eaxl xG &7t6 xfjc; near)? xe- 
xpayGvt>y xav xo utio xGv dxptov Tiepie/ojjievov op'doyGviov 
I'aov f) xG duo xfjc; ^earjc; xexpayGvcp, ai xpeu; eiWteTai 
dvdXoyov eaovxai. 

A' 1 

B' 1 A i 1 

r> ' 

"Eaxwaav xpeu; eu-deTai dvdXoyov ai A, B, r, Gc; f] A 
Tipoc; xf]v B, ouxtoc; rj B Tipoc; xf]v E Xeyco, oxi xo Oreo xGv 

A, r Tiepie)(6^evov op-doyGviov I'aov eaxl xG diio xfjc; B 
xexpayGvtp. 

Eeia-dco xfj B iar] f] A. 

Kdi ercei eaxiv Gc; f] A Tipoc; xf|V B, ouxw<; f] B Tipoc; xf]v 
T, iar) 8e f] B xfj A, eaxiv dpa Gc; f] A Tipoc; xrjv B, f] A Tipoc; 
xf|v r. edv 8e xeaaapec; euiDeTai dvdXoyov Gaiv, xo utio twv 
dxpwv Tiepie)(6^evov [opiJoyGviov] i'aov eaxl xo uko xGv 
(irotov Tiepiexo^iEvo opiJoyoviw. xo dpa utio tuv A, T iaov 
eaxl xG utio xGv B, A. dXXd xo utio xGv B, A xo duo xfjc; B 
eaxiv iar] ydp r\ B xfj A- xo dpa utio twv A, T Tiepie)(6[ievov 
opiJoyGviov laov eaxl xo duo xfjc; B xexpayGvw. 

AXXd 8f] xo utio xGv A, T iaov eaxo xG duo xfj? B' 
Xeyo, oxi eaxiv Gc; f) A Tipoc; xrjv B, ouxwc; f] B Tipoc; xf|V T. 

TGv yap auxGv xaxaaxeuaai!)evxwv, euel xo utio xGv A, 
r I'aov eaxl xG duo xfjc; B, dXXd xo drco xfjc; B xo utio xGv 

B, A eaxiv Tor) yap fj B xfj A- xo dpa utio xGv A, T Taov 
eaxl xG utio xGv B, A. eav 8e xo utio xGv axpwv i'aov fj xG 
utio xGv ^teawv, ai xeaaapec; eu$eTai dvdXoyov eiaiv. eaxiv 
dpa G? f] A Tipoc; xfjv B, ouxw<; f] A Tipoc; xf)v T. i'ar) 8e f] B 
xfj A- Gc; dpa f] A Tipoc; xrjv B, ouxok f] B Tipoc; xf)v T. 

'Eav dpa xpelc; euiDeTai dvdXoyov Gaiv, xo utio xGv axptov 
Tiepiex6(ievov op-doyGviov Taov eaxl xG duo xfjc \xsor\z xe- 
xpayGvcp- xav xo utio xGv axpwv Tiepiexo^tevov op'doyGviov 
Taov fj xG aTio xfj? [Lear]z xexpayGvo, ai xpeu; eu-delai 
dvdXoyov eaovxai' ouep eSei 8eTc;ai. 



the rectangle contained by the middle (two) then the four 
straight-lines will be proportional. (Which is) the very 
thing it was required to show. 

Proposition 17 

If three straight-lines are proportional then the rect- 
angle contained by the (two) outermost is equal to the 
square on the middle (one). And if the rectangle con- 
tained by the (two) outermost is equal to the square on 
the middle (one) then the three straight-lines will be pro- 
portional. 

A i 1 

B i 1 D 1 

C' 1 

Let A, B and C be three proportional straight-lines, 
(such that) as A (is) to B, so B (is) to C. I say that the 
rectangle contained by A and C is equal to the square on 
B. 

Let D be made equal to B [Prop. 1.3]. 

And since as A is to B, so B (is) to C, and B (is) 
equal to D, thus as A is to B, (so) D (is) to C. And if 
four straight-lines are proportional then the [rectangle] 
contained by the (two) outermost is equal to the rectan- 
gle contained by the middle (two) [Prop. 6.16]. Thus, 
the (rectangle contained) by A and C is equal to the 
(rectangle contained) by B and D. But, the (rectangle 
contained) by B and D is the (square) on B. For B (is) 
equal to D. Thus, the rectangle contained by A and C is 
equal to the square on B. 

And so, let the (rectangle contained) by A and C be 
equal to the (square) onB. I say that as A is to B, so B 
(is) to C. 

For, with the same construction, since the (rectangle 
contained) by A and C is equal to the (square) on B. 
But, the (square) on B is the (rectangle contained) by B 
and D. For B (is) equal to D. The (rectangle contained) 
by A and C is thus equal to the (rectangle contained) by 
B and D. And if the (rectangle contained) by the (two) 
outermost is equal to the (rectangle contained) by the 
middle (two) then the four straight-lines are proportional 
[Prop. 6.16]. Thus, as A is to B, so D (is) to C. And B 
(is) equal to D. Thus, as A (is) to B, so B (is) to C. 

Thus, if three straight-lines are proportional then the 
rectangle contained by the (two) outermost is equal to 
the square on the middle (one). And if the rectangle con- 
tained by the (two) outermost is equal to the square on 
the middle (one) then the three straight-lines will be pro- 
portional. (Which is) the very thing it was required to 



173 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 




TCaxco f] [iev So-deTaa eui5ela f) AB, to 8e SoiSev 
eui9uypa[4iov to TE- 5eT Sf] duo xfj? AB eui!)e(a<; xG TE 
eui9uypd[4io o\ioi6v xe xal 6fio[«<; xel^ievov euiJuypaii^ov 
dvaypdijiai. 

'ETieCeu)(T5w r) AZ, xal auveaxdxo Tip6<; xfj AB eu'dela 
xal xdig Ttpoc; auxfj aruieioic; xou; A, B xfj |iev upoc; xG T 
ycovia iar] f) Otto HAB, xfj Be uno IAZ lar\ f] hub ABH. XoiTif] 
dpa f) utio rZA xfj utio AHB eaxiv iar) - laoyGviov dpa eaxl 
xo ZrA xpiytovov xG HAB xpiyGvw. dvdXoyov dpa eaxlv 
(be; f) ZA Ttpog xiqv HB, ouxcoc f] Zr Tcpog xf]v HA, xal f| 
TA Tcpoc; xf]v AB. TidXiv auveaxdxco npog xfj BH eui5eia xal 
xou; Tcpoc; auxfj arpsioiz xolc; B, H xrj \ie\i utco AZE ywvia 
lot] f] utio BH0, xfj Se utio ZAE Tar) f] Otto HBO. Xoinf) dpa 
f) Ttpoc; xG E XoiTtfj xfj Ttpoc; xG eaxiv Iar)' laoyGviov dpa 
eaxl xo ZAE xpiyiovov xG HOB xpiyGvcy dvdXoyov dpa 
eaxlv Gc; f) ZA Ttp6<; xfjv HB, ouxcog f) ZE Tcpoc; xfjv HO xal 
f) EA Tipog xfjv OB. sSei/iSr) 8e xal Gc; f) ZA Tcpo<; xf)v HB, 
ouxcoc r) Zr Tcpoc; xf)v HA xal f) TA Ttpoc; xf)v AB- xal Gc; 
dpa f) Zr Ttpoc; xf)v AH, ouxmc; fj xe TA Ttpoc; xfjv AB xal f] 
ZE Ttpoc; xf)v HO xal exi f) EA Ttpoc; xf)v OB. xal eTtel Iar) 
eaxlv f) [xev Gito TZA y«v[a xfj utio AHB, r) 8e bub AZE xfj 
utio BHO, 6Xr) dpa f) utio TZE 6Xr) xfj utio AHO eaxiv Iar). 
6ia xa auxd 6f) xal f] utio IAE xfj Otto ABO eaxiv iar]. eaxi 
8e xal f) [iev Ttpoc; xG T xfj Ttpoc; xG A I'ar), f) 8s Ttpoc; xG E 
xfj Ttpoc; xG O. laoyGviov dpa eaxl xo AO xG TE- xal xac; 
itepl xac; laac; ycoviac; aG xGv TtXeupdc; dvdXoyov e/ei- o[ioiov 
dpa eaxl xo AO eu^uypajujiov xG TE eMuypd^cp. 

Atio xrjc; SoiSeia/jt; dpa cu^eiac; xfjg AB xG Scdevxi 
eu'duypdfi^cp xG TE 6y.oi6\ xe xal by.oioic, xei[ie\ov eu-duypa- 
y.[io\ dvayeypanxai xo AO- orcep eSei rcoifjaai. 



show. 

Proposition 18 

To describe a rectilinear figure similar, and simi- 
larly laid down, to a given rectilinear figure on a given 
straight-line. 




Let AB be the given straight-line, and CE the given 
rectilinear figure. So it is required to describe a rectilinear 
figure similar, and similarly laid down, to the rectilinear 
figure CE on the straight-line AB. 

Let DF have been joined, and let GAB, equal to the 
angle at C, and ABG, equal to (angle) CDF, have been 
constructed on the straight-line AB at the points A and 
B on it (respectively) [Prop. 1.23]. Thus, the remain- 
ing (angle) CFD is equal to AGB [Prop. 1.32]. Thus, 
triangle FCD is equiangular to triangle GAB. Thus, 
proportionally, as FD is to GB, so FC (is) to GA, and 
CD to AB [Prop. 6.4]. Again, let BGH, equal to an- 
gle DFE, and GBH equal to (angle) FD£, have been 
constructed on the straight-line BG at the points G and 
B on it (respectively) [Prop. 1.23]. Thus, the remain- 
ing (angle) at E is equal to the remaining (angle) at H 
[Prop. 1.32]. Thus, triangle FDE is equiangular to tri- 
angle GHB. Thus, proportionally, as FD is to GB, so 
FE (is) to GB, and ED to HB [Prop. 6.4]. And it was 
also shown (that) as FD (is) to GB, so FC (is) to GA, 
and CD to AB. Thus, also, as FC (is) to ^G, so CD (is) 
to AB, and FE to Gi7, and, further, ED to i/B. And 
since angle CFD is equal to AGB, and DF£ to BCff, 
thus the whole (angle) CFE is equal to the whole (an- 
gle) AGH. So, for the same (reasons), (angle) CDE is 
also equal to ABH. And the (angle) at C is also equal 
to the (angle) at A, and the (angle) at E to the (angle) 
at H. Thus, (figure) AH is equiangular to CE. And (the 
two figures) have the sides about their equal angles pro- 
portional. Thus, the rectilinear figure AH is similar to the 
rectilinear figure CE [Def. 6.1]. 

Thus, the rectilinear figure AH, similar, and similarly 
laid down, to the given rectilinear figure CE has been 
constructed on the given straight-line AB. (Which is) the 



174 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



I'd'. 

Ta o^ioia xpiycova npoc; dXXrjXa ev SmXaoiovi Xoycp eaxi 
xtbv o^oXoycov nXsupcbv. 

A 




B H T E Z 

'Eaxco ojioia xpiycova xd ABr, AEZ larjv £)(ovxa xrjv 
npog xcp B ycoviav xfj npoc; xcp E, coc Se xrjv AB npoc; xrjv Br, 
ouxcoc; xrjv AE npoc; xrjv EZ, coaxe 6[i6Xoyov sTvai xrjv Br 
xfj EZ- Xeyco, oxi xo ABr xpiycovov npoc; xo AEZ xpiycovov 
8inXaaiova Xoyov ex £1 *Fep ^1 Br npog xrjv EZ. 

EiXrjcp'dco yap xcov Br, EZ xpixrj dvdXoyov fj BH, coaxe 
elvai <b<; xrjv Br npoc; xrjv EZ, ouxcoc; xrjv EZ npoc; xrjv BH- 
xal STie^su/'dw rj AH. 

'End ouv eaxiv cbc; rj AB npoc; xf)v Br, ouxcoc; f] AE npoc; 
xrjv EZ, svaXXac; apa eaxlv coc; f) AB npoc; xrjv AE, ouxcoc; f] 
Br npoc; xrjv EZ. dXX' cbc; rj BT npoc EZ, ouxcoc; eaxiv f) EZ 
npoc; BH. xal cbc; apa fj AB npoc; AE, ouxcoc; fj EZ npoc; BH- 
xcov ABH, AEZ apa xpiycbvcov dvxmsnovdaaiv al nXsupal ai 
nepi xac; loac; ycovdic;. cov Se ^iav [iia !arjv exovxeov ycoviav 
xpiycbvcov dvxinenovdaoiv ai nXeupal ai tie pi xac; Iaac; ycovdic;, 
laa eaxlv exelva. ioov apa eaxi xo ABH xpiycovov xcp AEZ 
xpiycbvcp. xal enei eaxiv cbc; rj Br npoc xrjv EZ, ouxcoc; rj 
EZ npoc; xrjv BH, eav 8e xpelc; eu$eTai dvdXoyov coaiv, fj 
npcoxrj npoc; xrjv xpixrjv 8mXaa[ova Xoyov e)(ei f]TC£p npoc; 
xrjv Seuxepav, f) Br apa npoc; xrjv BH 8mXaaiova Xoyov 
£X£i f^Tcep fj TB npoc; xrjv EZ. cbc; 8s f] TB npoc; xrjv BH, 
ouxcoc; xo ABr xpiycovov npoc; xo ABH xpiycovov xal xo 
ABr apa xpiycovov npoc; xo ABH 8mXaa[ova Xoyov e)(ei 
rjnep f\ Br npoc; xrjv EZ. Ioov 8e xo ABH xpiycovov xcp 
AEZ xpiycbvcp. xal xo ABr apa xpiycovov npoc; xo AEZ 
xpiycovov 8mXaaiova Xoyov eys\. rjnep fj Br npoc "crjv EZ. 

Ta apa o^toia xpiycova npoc; aXXrjXa ev 8mXaoiovi Xoyco 
eaxi xcov o^ioXoycov nXeupcbv. [onep eSei 8ac;ai.] 



liopiajjia. 

'Ex 6rj xouxou cpavepov, oxi, eav xpeu; eO'delai dvdXoyov 
coaiv, eaxiv cbc; fj npcbxrj npoc; xrjv xpixrjv, ouxcoc; xo dno 



very thing it was required to do. 

Proposition 19 



Similar triangles are to one another in the squared* 
ratio of (their) corresponding sides. 
A 




B G C E F 



Let ABC and DEF be similar triangles having the 
angle at B equal to the (angle) at E, and AB to BC, as 
DE (is) to EF, such that BC corresponds to EF. I say 
that triangle ABC has a squared ratio to triangle DEF 
with respect to (that side) BC (has) to EF. 

For let a third (straight-line), BG, have been taken 
(which is) proportional to BC and EF, so that as BC 
(is) to EF, so EF (is) to BG [Prop. 6.11]. And let AG 
have been joined. 

Therefore, since as AB is to BC, so DE (is) to EF, 
thus, alternately, as AB is to DE, so BC (is) to EF 
[Prop. 5.16]. But, as BC (is) to EF, so £F is to BG. 
And, thus, as (is) to DE, so SB (is) to BG. Thus, 
for triangles ABC and DEF, the sides about the equal 
angles are reciprocally proportional. And those triangles 
having one (angle) equal to one (angle) for which the 
sides about the equal angles are reciprocally proportional 
are equal [Prop. 6.15]. Thus, triangle ABG is equal to 
triangle DEF. And since as BC (is) to EF, so EF (is) 
to BG, and if three straight-lines are proportional then 
the first has a squared ratio to the third with respect to 
the second [Def. 5.9], BC thus has a squared ratio to BG 
with respect to (that) CB (has) to EF. And as CB (is) 
to BG, so triangle ABC (is) to triangle ABG [Prop. 6.1]. 
Thus, triangle ABC also has a squared ratio to (triangle) 
ABG with respect to (that side) BC (has) to EF. And 
triangle ABG (is) equal to triangle DEF. Thus, trian- 
gle ABC also has a squared ratio to triangle DEF with 
respect to (that side) BC (has) to EF. 

Thus, similar triangles are to one another in the 
squared ratio of (their) corresponding sides. [(Which 
is) the very thing it was required to show] . 

Corollary 

So it is clear, from this, that if three straight-lines are 
proportional, then as the first is to the third, so the figure 



175 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



xrjc 7ipwxr](; sTSog Tipoc; xo anb xrjc; Seuxepac; xo 6[ioiov xal 
ojiolwc; dvaypacpo^ievov. oxerp eSei 8eic;ai. 

t Literally, "double". 

X . 

Td o\Loia TtoXuycova sic; xs o^oia xpiyiova 8iaipeTxai xal 
etc; laa xo TiXrj'doc; xal 6(i6Xoya xou; oXoic;, xal xo TtoXuycovov 
Tipoc; xo xoXuycovov SiTiXaaiova Xoyov *F £ P ^ o^ioXoyoc; 
xXeupd Tipoc; xr)v o^ioXoyov TtXsupdv. 




r a 

"Eaxw ojioia TtoXuycova xd ABrAE, ZHOKA, o^ioXoyoc; 
8e eox« f] AB xrj ZH- Xsyw, oxi xd ABrAE, ZH0KA 
TtoXuytova eic; xe o^ioia xpiywva BiaipeTxai xal sic; 'laa xo 
TiXrj'doc; xal 6(i6Xoya xolc; oXoic;, xal xo ABrAE TioXuyiovov 
Tipoc; xo ZHOKA xoXuycovov BiTtXaaiova Xoyov e);ei rjTtep f] 
AB Tipoc; xr]v ZH. 

Tks^eux^aav ai BE, EI\ HA, A6. 

Kal ind o^ioiov saxi xo ABrAE TioXuycovov xfi 
ZHOKA TcoXuywva), Tar) eraxlv r) utio BAE yiovia xfj utco 
HZA. xai saxiv tbc; f] BA Tipoc; AE, ouxwc; f] HZ Tipoc; ZA. 
end ouv 860 xpiywvd eaxi xd ABE, ZHA (Jiav ywvlav [iia 
ycovia ia/]v £)(ovxa, Tispl 8e xdc; I'aac; yioviac; xdc; TiXeupdc; 
dvdXoyov, laoycjviov dpa eaxl xo ABE xpiycovov xfi ZHA 
xpiycovcp- &>ots xal o^oiov lor) dpa sraxlv r) 0x6 ABE ycovia 
xfj utio ZHA. saxi 8e xal oXrj f] utio ABr oXrj xrj utio 
ZHO lat] 8id xr]v 6[ioioxr)xa xcov TioXuytovtov Xoixr] dpa f] 
utio EBr y«v£a xfj utio AH© eaxiv iar\. xal etcsl 8id xrjv 
6|jioL6xr)xa xwv ABE, ZHA xpiywvtov saxlv (I>c; rj EB Tipoc; 
BA, ouxmcj f) AH Ttpocj HZ, dXXd [ir)\ xal Sid xr|v o^ioioxrjxa 
xfiv TcoXuycovwv saxlv tbcj f) AB Tipoc; Br, ouxwc; f) ZH Tipoc; 
HO, 81' igou dpa eaxlv «c; f) EB Tipoc; Br, ouxcoc; i) AH Tipoc; 
H9, xal Tiepl xdc; I'oac; ycovdic; xdc; utio EBr, AH0 ai xXeupal 
dvdXoyov eiaiv iooycbviov dpa eaxl xo EBr xpiyovov iw 
AH© xpiywvo" oaxe xal 6[ioi6v iaxi xo EBr xpiycovov 
xfi AH© xpiytovw. 81a xd auxd 8r) xal xo ErA xpiywvov 
o^ioiov eoxi ifi A0K xpiywvw. xd dpa o^ioia TtoXuycova xd 
ABrAE, ZHOKA eic; xe o|ioia xpiycova 8if]pr)xai xal eic; laa 



(described) on the first (is) to the similar, and similarly 
described, (figure) on the second. (Which is) the very 
thing it was required to show. 



Proposition 20 

Similar polygons can be divided into equal numbers 
of similar triangles corresponding (in proportion) to the 
wholes, and one polygon has to the (other) polygon a 
squared ratio with respect to (that) a corresponding side 
(has) to a corresponding side. 



A 




C D 

Let ABCDE and FGHKL be similar polygons, and 
let AB correspond to FG. I say that polygons ABCDE 
and FGHKL can be divided into equal numbers of simi- 
lar triangles corresponding (in proportion) to the wholes, 
and (that) polygon ABCDE has a squared ratio to poly- 
gon FGHKL with respect to that AB (has) to FG. 

Let BE, EC, GL, and LH have been joined. 

And since polygon ABCDE is similar to polygon 
FGHKL, angle BAE is equal to angle GFL, and as BA 
is to AE, so GF (is) to FL [Def. 6.1]. Therefore, since 
ABE and FGL are two triangles having one angle equal 
to one angle and the sides about the equal angles propor- 
tional, triangle ABE is thus equiangular to triangle FGL 
[Prop. 6.6]. Hence, (they are) also similar [Prop. 6.4, 
Def. 6.1]. Thus, angle ABE is equal to (angle) FGL. 
And the whole (angle) ABC is equal to the whole (angle) 
FGH, on account of the similarity of the polygons. Thus, 
the remaining angle EBC is equal to LGH. And since, 
on account of the similarity of triangles ABE and FGL, 
as EB is to BA, so LG (is) to GF, but also, on account of 
the similarity of the polygons, as AB is to BC, so FG (is) 
to GH, thus, via equality, as EB is to BC, so LG (is) to 
GH [Prop. 5.22], and the sides about the equal angles, 
EBC and LGH, are proportional. Thus, triangle EBC is 
equiangular to triangle LGH [Prop. 6.6]. Hence, triangle 
EBC is also similar to triangle LGH [Prop. 6.4, Def. 6.1]. 
So, for the same (reasons), triangle ECD is also similar 



176 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



to 'KXfj'dog. 

Aeyto, oxi xal 6[i6Xoya xou; oXoic;, xouxeaxiv &axe 
dvdXoyov elvai xd xpiywva, xal r]yo6[jieva ^tev elvai xd ABE, 
EBr, ErA, eTto^ieva 8e auxwv xd ZHA, AH9, A0K, xal 
oxi xo ABrAE TtoXuywvov xpbc, xo ZH0KA TtoXuycovov 
8iTtXaaiova Xoyov l/ei r^Ttep f) ojioXoyoc; TtXeupd Ttpoc; xr]v 
o^ioXoyov TtXeupdv, xouxeaxiv f] AB Ttpoc; x/]v ZH. 

, ETte£eu)( , dcoaav yap ai Ar, Z0. xal eitel Bid x/]v 
6^oioxr]xa xwv TtoXuycovcov iar] eaxlv f] Otco ABr ycovia xfj 
uno ZHO, xai eaxiv £><; f] AB Ttpoc; Br, ouxwc; f) ZH Ttpoc; 
H9, laoytoviov eaxi xo ABr xpiywvov xw ZH0 xpiywvar 
tar) dpa eaxlv r] ^xev bub BAT ywvia xfj UTto HZ0, f) 8e 0ti6 
BEA xfj UTto H0Z. xal CTtel larj eaxlv r] 0ti6 BAM yovia 
xfj UTto HZN, eaxi 8e xal f] bub ABM xfj UTto ZHN iar), 
xal Xomf] dpa f] UTto AMB XoiTtrj xrj UTto ZNH iar) eaxiv 
laoywviov dpa eaxi xo ABM xplywvov xw ZHN xpiyova). 
o^ioiwc; 8rj 8eT^o[iev, 6x1 xal xo BMr xpiywvov laoywviov 
eaxi xw HN0 xpiycovw. dvdXoyov dpa eaxiv, cbc; y.sv rj AM 
Ttpoc; MB, ouxcoc; f) ZN Ttpoc; NH, cbc; 8e f] BM Ttpoc; Mr, 
ouxoc; f) HN Ttpoc; N0- cbaxe xal 81'' I'aou, cbc; f) AM Ttpoc; 
Mr, ouxwc; f] ZN Ttpoc; N6. dXX' <b<; f] AM Ttpoc Mr, ouxwc; 
xo ABM [xpiycovov] Ttpoc; xo MBr, xal xo AME Ttpoc; xo 
EME Ttpoc; dXXrjXa ydp eiaiv cbc; ai paaeu;. xal cbc; dpa 
ev ifiv rjyou^ivwv Ttpoc; ev tuv CTtou-evcov, ouxoc; aTtavxa 
xd f]you^teva Ttpoc; aTtavxa xd CTtojieva' cbc; dpa xo AMB 
xpiywvov Ttpoc; xo BMr, ouxwc; xo ABE Ttpoc; xo TBE. aXX' 
cbc; xo AMB Ttpoc; xo BMr, ouxtoc; f) AM Ttpoc; Mr- xal 
cbc; dpa f\ AM Ttpoc; Mr, ouxcoc; xo ABE xpiywvov Ttpoc; xo 
EBr xplywvov. Sid xd auxd 8r] xal cbc; f) ZN Ttpoc; N0, 
ouxoc; xo ZHA xpiywvov Ttpoc; xo HAG xpiywvov. xai eaxiv 
(be; f] AM Tipoc Mr, ouxw<; f) ZN Ttpoc; N0- xal (be; dpa 
xo ABE xpiyovov Ttpoc; xo BEr xpiyuvov, ouxtoc; xo ZHA 
xpiywvov icpoc xo HA9 xpiywvov, xal evaXXd? &>c, xo ABE 
xpiyovov Ttp6<; xo ZHA xpiyovov, ouxoc xo BEr xpiywvov 
jcp6<; xo HA6 xpiywvov. ojioiclx; 8f) 8£[^o[iev eTciC£U)fdeiaa)v 
xwv BA, HK, oxi xal w<; xo BEr xpiywvov icpo^ xo AH0 
xpiywvov, ouxw<; xo ErA xpiywvov Ttpoc; xo A0K xptywvov. 
xal ETtst saxtv «<; xo ABE xptywvov Ttpoc xo ZHA xptywvov, 
oux6X xo EBr Ttpoc; xo AH0, xal exi xo ErA Ttpoc; xo A0K, 
xal (be; dpa ev xwv fjyou^fvwv Ttpoc; ev xwv CTtojievcov, ouxoc 
aTtavxa xd fpfou\ievoL Ttpoc; aTtavxa xd STto^ieva' eaxiv dpa 
&>Z xo ABE xpiyovov Ttpoc; xo ZHA xpiyovov, ouxwc; xo 
ABrAE TtoXuywvov Ttpoc; xo ZH6KA TtoXuywvov. dXXd xo 
ABE xpiywvov Ttpoc; xo ZHA xpiywvov SmXaabva Xoyov 
eX^i ^Ttep f) AB ojioXoyoc; TtXeupd Ttpoc; xr)v ZH o^ioXoyov 
TtXsupdv xd yap o^ioia xpiywva ev 8iTtXaa(ovi Xoyco eaxl 
xwv o^ioXoywv TtXeupwv. xal xo ABrAE dpa TtoXuywvov 
Ttpoc; xo ZH0KA TtoXuywvov 8iTtXaaiova Xoyov e)(ei f]Ttep rj 
AB ojioXoyoc; TtXeupd Ttpoc; xrjv ZH o^ioXoyov TtXeupdv. 

Td dpa o^ioia TtoXuywva eic; xe o^ioia xptyova Biaipelxai 
xal el? laa xo TtXfj'doc; xal ojioXoya xoTc; oXoic;, xal xo 



to triangle LHK. Thus, the similar polygons ABCDE 
and FGHKL have been divided into equal numbers of 
similar triangles. 

I also say that (the triangles) correspond (in propor- 
tion) to the wholes. That is to say, the triangles are 
proportional: ABE, EBC, and ECD are the leading 
(magnitudes), and their (associated) following (magni- 
tudes are) FGL, LGH, and LHK (respectively) . (I) also 
(say) that polygon ABCDE has a squared ratio to poly- 
gon FGHKL with respect to (that) a corresponding side 
(has) to a corresponding side — that is to say, (side) AB 
to FG. 

For let AC and FH have been joined. And since angle 
ABC is equal to FGH, and as AB is to BC, so FG (is) to 
GH, on account of the similarity of the polygons, triangle 
ABC is equiangular to triangle FGH [Prop. 6.6]. Thus, 
angle BAC is equal to GFH, and (angle) BCA to GHF. 
And since angle BAM is equal to GFN, and (angle) 
ABM is also equal to FGN (see earlier), the remaining 
(angle) AMB is thus also equal to the remaining (angle) 
FNG [Prop. 1.32]. Thus, triangle ABM is equiangular 
to triangle FGN. So, similarly, we can show that triangle 
BMC is also equiangular to triangle GNH. Thus, pro- 
portionally, as AM is to MB, so FN (is) to NG, and as 
BM (is) to MC, so GN (is) to NH [Prop. 6.4]. Hence, 
also, via equality, as AM (is) to MC, so FN (is) to A^i? 
[Prop. 5.22]. But, as AM (is) to MC, so [triangle] ABM 
is to MBC, and AME to £MC. For they are to one an- 
other as their bases [Prop. 6.1]. And as one of the leading 
(magnitudes) is to one of the following (magnitudes), so 
(the sum of) all the leading (magnitudes) is to (the sum 
of) all the following (magnitudes) [Prop. 5.12]. Thus, as 
triangle AMB (is) to BMC, so (triangle) ABE (is) to 
CBE. But, as (triangle) AMB (is) to BMC, so AM (is) 
to MC. Thus, also, as AM (is) to MC, so triangle ABE 
(is) to triangle EBC. And so, for the same (reasons), as 
FN (is) to NH, so triangle FGL (is) to triangle GLH. 
And as AM is to MC, so i^N (is) to NH. Thus, also, as 
triangle ABE (is) to triangle BEC, so triangle i^GL (is) 
to triangle GLH, and, alternately, as triangle ABE (is) 
to triangle FGL, so triangle BEC (is) to triangle GLi? 
[Prop. 5.16]. So, similarly, we can also show, by joining 
BD and GK, that as triangle BEC (is) to triangle LGH, 
so triangle ECD (is) to triangle LHK. And since as tri- 
angle ABE is to triangle FGL, so (triangle) £BG (is) 
to LGH, and, further, (triangle) ECD to LHK, and also 
as one of the leading (magnitudes is) to one of the fol- 
lowing, so (the sum of) all the leading (magnitudes is) to 
(the sum of) all the following [Prop. 5.12], thus as trian- 
gle ABE is to triangle FGL, so polygon ABCDE (is) to 
polygon FGHKL. But, triangle ABE has a squared ratio 



177 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



rcoXuyiovov rcpoc; to rcoXuywvov 8ircXaaiova Xoyov e)(ei f^TCEp 
f] 6[i6Xoyo<; rcXcupa rcpoc; ttjv ojioXoyov rcXcupdv [orcsp sSel 
8eT^ai] . 



I16pia[Jia. 

'Qaauxcoc; Se xal ski xfiiv [ojioiwv] xsxparcXsuptov Seix^- 
aexai, oxi ev BircXaafovi Xoyo elol xwv 6^oX6y«v rcXeupfiv. 
eBelx^t] 8e xal eni x£5v xpiycovcov wots xal xai&oXou xa 
ojKna eu-duypa^iia ax^liaxa rcpoc; aXXrjXa ev BircXaalovi 
Xoycp elol iSv ojioXoycov rcXeupwv. orcep eSei 8ei?ai. 

xa'. 

Ta iS auxfi eO'duypd^cp o^ioia xal dXXr]Xoic; eaxlv 
ojioia. 




"Eaxw yap sxdxspov xfiv A, B eO'duypd^cov iw T 
o^ioiov Xsy«, oxi xal xo A tw B eaxiv o^oiov. 

'Ercri yap o^ioiov eaxi xo A xa> T, laoycoviov xe eaxiv 
auxw xal xac; rcepl xac; laag ycoviac; rcXeupac; dvdXoyov ey^ei. 
rcdXiv, end ojioiov eaxi xo B x£5 T, laoycoviov xe eaxiv 
auxo xal xac; rcepl xac; Taac; ycoviac; rcXeupac; dvdXoyov ex £l - 
exdxepov apa xwv A, B xG T icoywviov xe eaxi xal xac; 
rcepl xac; taac; ywviac; rcXeupac; dvdXoyov s^ei [waxe xal xo 
A tu B laoycoviov xe eaxi xal xac; rcspl xac foat; ytoviac; 



to triangle FGL with respect to (that) the corresponding 
side AB (has) to the corresponding side FG. For, similar 
triangles are in the squared ratio of corresponding sides 
[Prop. 6.14]. Thus, polygon ABCDE also has a squared 
ratio to polygon FGHKL with respect to (that) the cor- 
responding side AB (has) to the corresponding side FG. 

Thus, similar polygons can be divided into equal num- 
bers of similar triangles corresponding (in proportion) to 
the wholes, and one polygon has to the (other) polygon a 
squared ratio with respect to (that) a corresponding side 
(has) to a corresponding side. [(Which is) the very thing 
it was required to show] . 

Corollary 

And, in the same manner, it can also be shown for 
[similar] quadrilaterals that they are in the squared ratio 
of (their) corresponding sides. And it was also shown for 
triangles. Hence, in general, similar rectilinear figures are 
also to one another in the squared ratio of (their) corre- 
sponding sides. (Which is) the very thing it was required 
to show. 

Proposition 21 

(Rectilinear figures) similar to the same rectilinear fig- 
ure are also similar to one another. 




Let each of the rectilinear figures A and B be similar 
to (the rectilinear figure) C. I say that A is also similar to 
B. 

For since A is similar to C, {A) is equiangular to (C), 
and has the sides about the equal angles proportional 
[Def. 6.1]. Again, since B is similar to C, (B) is equian- 
gular to (C), and has the sides about the equal angles 
proportional [Def. 6.1]. Thus, A and B are each equian- 
gular to C, and have the sides about the equal angles 



178 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



TtXeupdc; dvdXoyov exei]. o^ioiov dpa sail to A tw B- oTtep 
e8e:i 8eTc;ai. 



x(3'. 

'Edv xeaaapec; eu^eTai dvdXoyov SSmv, xal xd an auxfiv 
eGTJuypajji^a opioid xe xdi ojioicoc; dvayeypa^eva dvdXoyov 
eaxai- xdv xd dm' auxwv et/duypaii^ia opioid xe xal o^oico? 
dvayeypa^eva dvaXoyov rj, xdi auxdi ad £0156101 dvdXoyov 
eaovxai. 




A B r A 




n p 

"Eaxwaav xeaaapec; euiMai dvdXoYov ai AB, TA, EZ, 
H0, (be; f) AB Ttpoc; xr)v TA, ouxgk r| EZ Ttpoc; xrjv H6, xdi 
dvayeypdcpiSioaav a7t0 ^ v T " v AB, TA opioid xe xdi ojioiioc; 
xei^xeva eMuypajjijia xd KAB, ArA, aTto Se xov EZ, H9 
opioid xe xdi o^ioicoc; xei^ieva eMuypa^a xd MZ, N6- Xeyw, 
oxi eaxiv foe, xo KAB Ttpoc; xo ArA, ouxgjc; xo MZ Ttpoc; xo 
N6. 

EiXricp-dw yap x«v |iev AB, TA xpixr) dvdXoyov f] S, xfiv 
8e EZ, H9 xpixr) dvdXoyov f] 0. xdi enei eaxiv tbc; [iev f) AB 
Ttpoc; x/)v TA, ouxioc; f) EZ Ttpoc; xrjv H6, tbc; 8e f) TA Ttpoc; 
xr]v S, oux6K f] H6 Ttpoc; x/)v 0, 8i' laou dpa eaxiv (be; f) 
AB Ttpoc; xr)v S, ouxwg f) EZ Ttpoc; xrjv 0. dXX' tbc; jiev f] 
AB Ttpoc; x/)v S, ouxtoc; [xal] xo KAB Ttpoc; xo ArA, cbc; 8e 
f] EZ Ttpoc; x/)v 0, ouxwc; xo MZ Ttpoc; xo N6- xal (be; dpa 
xo KAB Ttpoc; xo ArA, ouxwc; xo MZ Ttpoc; xo N0. 

AXXd Sr) eaxto tbc; xo KAB Ttpoc; xo ArA, ouxwc; xo MZ 
Ttpoc; xo N9- Xeyw, oxi eaxl xal foe, rj AB Ttpoc; xrjv TA, 
ouxwc; f) EZ Ttpoc; xrjv H9. ei yap \ir\ eaxiv, foe, f) AB Ttpoc; 
xrjv TA, ouxwc; r) EZ Ttpoc; xrjv H9, eoxw (be; f] AB Ttpoc; xrjv 
TA, ouxoc r| EZ Ttpoc; xrjv IIP, xal dvayeypdcpiJo dito xrjc; 



proportional [hence, A is also equiangular to B, and has 
the sides about the equal angles proportional]. Thus, A 
is similar to B [Def. 6.1] . (Which is) the very thing it was 
required to show. 

Proposition 22 

If four straight-lines are proportional then similar, and 
similarly described, rectilinear figures (drawn) on them 
will also be proportional. And if similar, and similarly 
described, rectilinear figures (drawn) on them are pro- 
portional then the straight-lines themselves will also be 
proportional. 




A BCD 




Q R 

Let AB, CD, EF, and GH be four proportional 
straight-lines, (such that) as AB (is) to CD, so EF (is) 
to GH. And let the similar, and similarly laid out, rec- 
tilinear figures KAB and LCD have been described on 
AB and CD (respectively), and the similar, and similarly 
laid out, rectilinear figures MF and NH on EF and GH 
(respectively). I say that as KAB is to LCD, so MF (is) 
to NH. 

For let a third (straight-line) O have been taken 
(which is) proportional to AB and CD, and a third 
(straight-line) P proportional to EF and GH [Prop. 6.11]. 
And since as AB is to CD, so EF (is) to GH, and as CD 
(is) to 0, so GH (is) to P, thus, via equality, as AB is to 
0, so EF (is) to P [Prop. 5.22]. But, as AB (is) to 0, so 
[also] KAB (is) to LCD, and as EF (is) to P, so MF 
(is) to NH [Prop. 5.19 corr.]. And, thus, as KAB (is) to 
LCD, so MF (is) to NH. 

And so let KAB be to LCD, as MF (is) to NH. I say 
also that as AB is to CD, so EF (is) to Gi7. For if as AB 
is to CD, so £F (is) not to GH, let ylB be to CD, as £F 



179 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



IIP oiioxepcp xfiv MZ, N9 o^oiov xe xal o^ioicoc; xei|jievov 
eu'duypa^tuov to EP. 

'Ercel ouv eaxiv (be; f] AB 7ipo<; xrjv TA, ouxw<; f] EZ upoc; 
x/]v IIP, xal dvayeypaTixai a7l ° ^ v T " v AB, TA opioid xe 
xal 6(ioia>c xeifieva xd KAB, ABA, duo 8e xebv EZ, nP 
Sfioid xe xal o^otox xei^ieva xd MZ, SP, eaxiv apa cbc; xo 
KAB Tip6<; xo ArA, ouxclx; xo MZ npbc, xo SP. tmoxeixai 
8e xal <b<; xo KAB 7ipo<; xo ArA, ouxwc; xo MZ 7ipo<; xo 
N6- xal cbc; apa xo MZ Ttpoc. xo EP, ouxwe. xo MZ npbc, 
xo N0. xo MZ apa Ttpoc, exdxepov xwv N0, SP xov auxov 
§XCi Xoyov laov apa eaxl xo N0 tw SP. eaxi 8s auxw xal 
o^toiov xal b[io'ux>z xei^tevov I'ar) apa f] H9 xrj nP. xal ernei 
eaxiv cb<; f] AB Ttpoc. xrjv TA, ouxck f) EZ Ttpoc. xrjv nP, iar] 
8s f] nP xrj H0, eaxiv apa (be. f) AB Tcpoc. xrjv PA, ouxoc. rj 
EZ Ttpoc, xr]v H9. 

'Edv apa xeaaapec, eu'delai dvdXoyov waiv, xal xd dm' 
auxCSv sO'duypaji^.a opioid xe xal 6^oio<; dvayeypa^eva 
dvdXoyov eaxar xav xd dm' auxebv eCcvMypamia opioid xe 
xal 6\ioi(x>q, dvayeypa^eva dvdXoYov fj, xal auxai al eu-fMai 
dvdXoYov eaovxai' onep eSei 8eTc;ai. 



(is) to QR [Prop. 6.12]. And let the rectilinear figure SR, 
similar, and similarly laid down, to either of MF or NH, 
have been described on QR [Props. 6.18, 6.21]. 

Therefore, since as AB is to CD, so EF (is) to QR, 
and the similar, and similarly laid out, (rectilinear fig- 
ures) KAB and LCD have been described on AB and 
CD (respectively), and the similar, and similarly laid out, 
(rectilinear figures) MF and SR on EF and QR (re- 
sespectively), thus as KAB is to LCD, so MF (is) to 
SR (see above). And it was also assumed that as KAB 
(is) to LCD, so MF (is) to NH. Thus, also, as MF (is) 
to SR, so MF (is) to NH [Prop. 5.11]. Thus, MF has 
the same ratio to each of NH and SR. Thus, AT? is equal 
to SR [Prop. 5.9]. And it is also similar, and similarly laid 
out, to it. Thus, GH (is) equal to QRJ And since AB is 
to CD, as EF (is) to QR, and QR (is) equal to GH, thus 
as AB is to CD, so EF (is) to Cff. 

Thus, if four straight-lines are proportional, then sim- 
ilar, and similarly described, rectilinear figures (drawn) 
on them will also be proportional. And if similar, and 
similarly described, rectilinear figures (drawn) on them 
are proportional then the straight-lines themselves will 
also be proportional. (Which is) the very thing it was 
required to show. 



t Here, Euclid assumes, without proof, that if two similar figures are equal then any pair of corresponding sides is also equal. 



xy'. 

Td iaoy(bvia TiapaXXr]X6Ypajj.(jia npbc; dXXrjXa Xoyov e)(ei 
xov auyxei^ievov ex xGv TiXeupcbv. 

'Eaxto iaoy(bvia TtapaXXr)X6ypa|ji[j.a xd Ar, TZ Tarjv 
E^ovxa xrjv utio BTA ywviav xrj utio EITT Xeycj, oxi xo Ar 
TiapaXXr)X6ypajj.|jiov Ttpoc, xo TZ TtapaXXr)X6ypa(i|j.ov Xoyov 
e/ei xov auyxeijievov ex x(5v TtXeupfiiv. 

Kslcrdco yap (baxe in cu^eiac, elvai xr)v Br xrj TH- in 
eMeiac. apa eaxl xal f] AT xrj TE. xal au[iTteTtXY]p(baTL>cd xo 
AH 7iapaXXr]X6ypa[ji[j.ov, xal exxeicrdGj xic, cu-dela f] K, xal 
yeyovexw (be. \xe\ f] BT Ttpoc, xy]v TH, ouxmc. f) K npbc, xrjv 
A, (be 8s f] AT npbc, x/)v TE, ouxioc; f) A npog x/)v M. 

Oi apa Xoyoi xfjg xs K upoc; xrjv A xal xrjc; A upoc 
xr)v M oi auxoi eiai zoic, Xoyoic; x«v uXsupcbv, xfjc ^£ BT 
Tipog xrjv TH xal xrjg Ar Ttpoc; xr)v TE. dXX' 6 xfjc; K Tipoc 
M Xbyoc, auyxeixai ex xe xou jy]z K up6(; A Xoyou xal 
xou xfj<; A 7ip6<; M- tbaxe xal f) K 7tpo<; xf]v M Xoyov exei 
xov auyxel^evov ex xebv nXeupfiv. xal enei eaxiv cbi; f) Br 
Tipog xr]v TH, ouxtog xo Ar TiapaXXrjXoypa^^tov Tip6(; xo 
TQ, dXX' <bg f) BT rcpoc; xf)v TH, ouxoc; f) K upog xrjv A, 
xal (b<; apa f) K Ttpoc; xt]v A, oux«c xo AT npbz xo T6. 
TidXiv, eitei eaxiv ok f\ AT npbz xr]v TE, ouxwc; xo TQ ita- 
paXX/]X6ypa^i[iov npbc, xo TZ, dXX' cbg f] Ar 7tp6<; x/]v TE, 



Proposition 23 

Equiangular parallelograms have to one another the 
ratio compounded^ out of (the ratios of) their sides. 

Let AC and CF be equiangular parallelograms having 
angle BCD equal to ECG. I say that parallelogram AC 
has to parallelogram CF the ratio compounded out of 
(the ratios of) their sides. 

For let BC be laid down so as to be straight-on to 
CG. Thus, DC is also straight-on to CE [Prop. 1.14]. 
And let the parallelogram DG have been completed. And 
let some straight-line K have been laid down. And let it 
be contrived that as BC (is) to CG, so K (is) to L, and 
as DC (is) to CE, so L (is) to M [Prop. 6.12]. 

Thus, the ratios of K to L and of L to M are the same 
as the ratios of the sides, (namely), BC to CG and DC 
to CE (respectively). But, the ratio of K to M is com- 
pounded out of the ratio of K to L and (the ratio) of L 
to M. Hence, K also has to M the ratio compounded 
out of (the ratios of) the sides (of the parallelograms). 
And since as BC is to CG, so parallelogram AC (is) to 
GH [Prop. 6.1], but as BC (is) to CG, so K (is) to L, 
thus, also, as K (is) to L, so (parallelogram) AC (is) to 
GH. Again, since as DC (is) to CE, so parallelogram 



180 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



outwc; f] A 7ip6<; x/]v M, xdx &>c, dpa f] A npbc, ttjv M, outox 
to T9 TtapaXX/jXoYpa^ov Tcpoc to TZ 7tapaXXr)X6Ypa^ov. 
etieI ouv eBeix'dr), cbc; uev f] K npog ttjv A, outck to Ar 
7iocpaXXr]X6Ypoc^ov 7ip6<; to TO 7tapaXX/]XoYpa^ov, w<; Be: 
f) A TCpog t/)v M, ouTtx; to T0 7iapaXX/]XoYpa|ji[j.ov upoc; to 
TZ 7tapaXXr]X6Ypa^[iov, 81' Taou apa eotIv cbc; f) K Ttpoc; tt]v 
M, outmc; to Ar Ttpoc; to TZ TtapaXXrjXoYpa^ov. f] 8e K 
7ip6<; Tf]v M Xoyov g)(ei tov auyxdjisvov ex t£>v uXsupwv 
xal to Ar apa icpoc; to TZ Xbyo\ s/ei tov a\jyxei[ievov tx 
iwv uXeupOv. 



CF (is) to CF [Prop. 6.1], but as DC (is) to CE, so L 
(is) to M, thus, also, as £ (is) to M, so parallelogram 
CH (is) to parallelogram CF. Therefore, since it was 
shown that as K (is) to L, so parallelogram AC (is) to 
parallelogram CH, and as L (is) to M, so parallelogram 
CH (is) to parallelogram CF, thus, via equality, as K is 
to M, so (parallelogram) AC (is) to parallelogram CF 
[Prop. 5.22]. And K has to M the ratio compounded out 
of (the ratios of) the sides (of the parallelograms) . Thus, 
(parallelogram) AC also has to (parallelogram) CF the 
ratio compounded out of (the ratio of) their sides. 







D 



H 



K' ' 

A' ' 

Mi ' j 

E Z 

Ta apa laoYiovia TcapaXX/]XoYpa[i^.a Tipoc; aXX/]Xa Xoyov 
£/si tov auYxeijievov ex x&m uXeupfiv ouep e8ei 8eT<;ai. 



B 

K ' 1 

L ' ' 

Mi 1 

E F 

Thus, equiangular parallelograms have to one another 
the ratio compounded out of (the ratio of) their sides. 
(Which is) the very thing it was required to show. 



t In modern terminology, if two ratios are "compounded" then they are multiplied together. 



x5'. 

TlavToc; 7iapaXX/]XoYpa[i^ou Ta itep! xr)v Sid^ieTpov ua- 
paXXrjXoYpa^a opioid taxi x& ts 6X« xal dXXr|Xou;. 

"EaTW TiapaXXrjXoYpa^i^ov to ABTA, 8id^£Tpoc; 8e 
auToO f) AT, Ttepl 8s ttjv AT TcapaXXrjXoYpa^a eaTW Ta EH, 
6K- Xeyw, oti cxaTspov t£Sv EH, 0K TtapaXXr)XoYpd^«v 
5\ioi6v taxi SXfc) to ABTA xal dXXfiXou;. 

'Etc! yap TpiY^vou tou ABT Ttapa (jiiav xSv icXeupwv 
t/]v BT rjxTai f] EZ, dvdXoYov icmv cbc; f] BE icpoc; t/]v 
EA, outca; f) TZ Ttpoc" ttjv ZA. icdXiv, etcI TpiYtbvou tou 
ATA icapa ^i(av ttjv TA rjxTai f\ ZH, dvdXoYov sgtiv ok f\ 
TZ npbc, tt]v ZA, outcx f] AH Ttpoc" ttjv HA. dXX'' cbc; r) 
TZ Ttpoc; t/]v ZA, outck eBdx'dr) xal r) BE Ttpoc; t/]v EA- 
xal cbc; apa f] BE Ttpoc; xf\v EA, outcoc; f] AH Tipoc; xf\v 
HA, xal auvdfvTi apa cbc; f] BA Ttpoc; AE, outcx r) AA 
Ttpoc; AH, xal evaXXac; cbc; r) BA Ttpoc; ttjv AA, outcoc; f) 
EA Ttpoc; tt]v AH. tc5v apa ABTA, EH TtapaXXr)XoYpd^«v 
dvdXoYov eiaiv ai TtXeupal al Ttepl Trjv xoiv/]v Y^viav ttjv utco 
BAA. xal sttei 7iapdXXr)X6(; ecrciv 1] HZ Tfj AT, lay] eotIv 
f] ^iev U7i6 AZH yav'ia xt] bub ATA- xal xoivr) xuv 8uo 



Proposition 24 

In any parallelogram the parallelograms about the di- 
agonal are similar to the whole, and to one another. 

Let ABCD be a parallelogram, and AC its diagonal. 
And let EG and HK be parallelograms about AC. I say 
that the parallelograms EG and HK are each similar to 
the whole (parallelogram) ABCD, and to one another. 

For since EF has been drawn parallel to one of the 
sides BC of triangle ABC, proportionally, as BE is to 
EA, so CF (is) to FA [Prop. 6.2]. Again, since FG has 
been drawn parallel to one (of the sides) CD of trian- 
gle ACD, proportionally, as CF is to FA, so DG (is) to 
GA [Prop. 6.2]. But, as CF (is) to FA, so it was also 
shown (is) BE to EA. And thus as f?F (is) to EA, so 
DG (is) to GA. And, thus, compounding, as BA (is) to 
AE, so DA (is) to AG [Prop. 5.18]. And, alternately, as 
BA (is) to AD, so EA (is) to AG [Prop. 5.16]. Thus, 
in parallelograms ABCD and EG the sides about the 
common angle BAD are proportional. And since GF is 
parallel to DC, angle AFC is equal to DGA [Prop. 1.29]. 



181 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



xpiyovwv xfiv AAr, AHZ f] uko A AT ywvla' laoywviov 
apa taxi xo AAr xpiycovov xfi AHZ xpiywvw. 8id xa auxa 
8r) xal xo ArB xpiywvov laoycoviov saxi iu AZE xpiyovw, 
xal oXov xo ABrA 7iapaXXr]X6ypa^ov iu EH TtapaXXr]- 
Xoypd^up laoycoviov saxiv. dvdXoyov apa laxlv cot; f) AA 
7ip6<; xrjv Ar, oux«<; f] AH 7tp6<; xr)v HZ, to? Ss f] Ar npoz 
xr]v TA, oOxto? f] HZ 7tp6<; xrjv ZA, d>c 8s f] Ar upot; x/]v 
TB, ouxw<; f] AZ Ttp6<; xrjv ZE, xal sxi cb<; f] TB 7tp6<; xrjv 
BA, oux«<; f] ZE Tipoc; xrjv EA. xal sitsl eSsix^ t^v 
f] Ar Ttpoc; xrjv TA, ouxw<; r\ HZ Ttpoc; xrjv ZA, £><; 8s f) 
Ar npd<z ir\\> TB, ouxwg f] AZ 7tp6<; xrjv ZE, 8i' laou apa 
saxlv cb<; r) Ar 7ip6<; xrjv TB, ouxtoc; f) HZ Ttp6<; xrjv ZE. 
xfiiv apa ABrA, EH 7iapaXXr)Xoypd^.[i«v dvaXoyov siaiv 
ai uXsupal ai nspi xdc; laac ycoviac ojioiov apa saxl xo 
ABrA 7iapaXXr]Xoypd^ov xw EH 7tapaXXr]Xoypd^tp. 8id 
xd auxa 8r) xo ABrA 7tapaXXr]X6ypa(i[jiov xal xw KO rca- 
paXXr)Xoypd^i[iw ojioiov saxiv sxdxspov apa xwv EH, 8K 
7tapaXXr]Xoypd^tov iu ABrA [n:apaXXr)Xoypd^w] o^ioiov 
saxiv. xa 8s xw auxw suTJuypd^tp o^ioia xal dXXrjXoic; saxlv 
ofjioia' xal xo EH apa TtapaXXr)X6ypa^ov x£> 0K TtapaXXrj- 
Xoypd[i^tw o\±oio\i saxiv. 



A E B 




A K r 

navxog apa TtapaXXr)Xoypd^ou xd uspl xiqv 8id^isxpov 
TtapaXXiqXoypa^a ojioid saxi xai xs oXo xal dXXrjXoic oTtsp 
s8si SsT^ai. 

xs'. 

Tcp SotJevxi su$uypd[i[jiw 6|ioiov xal aXXtp iw So-dsvxi 
I'aov xo auxo auaxr]aaa , dai. 



And angle DAC (is) common to the two triangles ADC 
and AGF. Thus, triangle ADC is equiangular to triangle 
AGF [Prop. 1.32]. So, for the same (reasons), triangle 
ACB is equiangular to triangle AFE, and the whole par- 
allelogram ABCD is equiangular to parallelogram EG. 
Thus, proportionally, as AD (is) to DC, so AG (is) to 
GF, and as DC (is) to CA, so GF (is) to FA, and as AC 
(is) to CB, so AF (is) to FF, and, further, as CB (is) 
to BA, so F F (is) to EA [Prop. 6.4]. And since it was 
shown that as DC is to CA, so GF (is) to FA, and as 
AC (is) to CB, so (is) to FE, thus, via equality, as 
DC is to CF, so GF (is) to FF [Prop. 5.22]. Thus, in 
parallelograms ABCD and EG the sides about the equal 
angles are proportional. Thus, parallelogram ABCD is 
similar to parallelogram EG [Def. 6.1]. So, for the same 
(reasons), parallelogram ABCD is also similar to par- 
allelogram KH. Thus, parallelograms EG and HK are 
each similar to [parallelogram] ABCD. And (rectilin- 
ear figures) similar to the same rectilinear figure are also 
similar to one another [Prop. 6.21]. Thus, parallelogram 
EG is also similar to parallelogram HK. 



A E B 




D K C 

Thus, in any parallelogram the parallelograms about 
the diagonal are similar to the whole, and to one another. 
(Which is) the very thing it was required to show. 

Proposition 25 

To construct a single (rectilinear figure) similar to a 
given rectilinear figure, and equal to a different given rec- 
tilinear figure. 



182 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 




A E M 



'Egxm to ^t£v BoiJev £ui56ypa|ji[iov, S 8a ojiolov 
auaxrpacrdai, to ABr, G 8e 8eT I'aov, to A- Sa 8rj iu 
[lev ABr o^ioiov, iS 8e A laov to auxo ouaxr]oaa , dai. 

napapepXf]a , dw yap Ttapd [Lev xr)v Br xG ABr xpiycovw 
laov 7iapaXXr]X6Ypa^ov xo BE, Ttapd 8e xrjv TE iS A '(aov 
napaXXrjXoYpa^ov xo TM sv yojvia xfj Otto ZrE, fj eaxiv 
Tar) xfj uiio TBA. in' ed-Qsiac, dpa saxlv f) ^ev Br xfj TZ, f| 
8e AE x^ EM. xal eiXfjcp'dw xfiv Br, TZ y.eor) dvaXoyov f) 
H9, xal dvaysypdcp^Gj arco xfjc H9 xfi ABr ojiolov xe xal 
o^lolmc xdjisvov xo KH9. 

Kal STiei saxiv foe, i] BT npbc, xfjv H9, oux«<; r) H9 
upog xr]v rZ, sdv 8e xpelc; sMelai dvaXoyov fiaiv, eaxiv 
foe, f] upwxr] Tipoc; xf]v xpixrjv, ouxclx; xo arco xfj? Ttpt5xr]<; 
eT8o<; Ttpoc; xo duo xfjc 8euxepa<; xo 6[ioiov xal o^oiox dva- 
Ypacpouevov, eaxiv dpa ox rj Br Ttp6<; xf]v TZ, ouxgx xo 
ABr xpiywvov Ttp<x xo KH9 xp(y«vov. dXXd xal ox f] BT 
7ip6<; xf|V rZ, ouxox xo BE TiapaXX^Xoypa^ov Ttpcx xo EZ 
7iapaXXr]X6Ypa[i^ov. xal ox apa xo ABr xpiycovov Ttpoc xo 
KH9 xpiywvov, ouxox xo BE TtapaXXrjXoYpa^ov Ttpcx xo 
EZ TtapaXX/]XoYpa^ov evaXXdc; apa ox xo ABr xpiywvov 
Ttp<x xo BE TiapaXX^Xoypa^^ov, ouxox xo KH9 xpiYO)vov 
7ip6<; xo EZ 7iapaXXr]X6Ypa^ov. taov 8e xo ABr xpiYOJvov 
xo BE TtapaXX/]XoYpd^oy laov dpa xal xo KH9 xpiYWvov 
xfi> EZ n;apaXXr)XoYpd^[io. dXXd xo EZ TiapaXX/jXoYpa^ov 
tu A eaxiv Taov xal xo KH9 apa xw A eaxiv iaov. eaxi 
8e xo KH9 xal x« ABr ojioiov. 

Top apa BoiJevxi eui!)uYpd|i^oj xoj ABr opioiov xal aXXo) 
iG BoiJevxi iS A I'aov xo auxo auveaxaxai xo KH9- oTtep 
e8ei Ttoirjaai. 



Xf'. 

Eav duo TiapaXXrjXoYpd^ou TiapaXX/jXoYpa^iov dcpai- 
pedf 6[ioi6v xe x« 0X0) xal opioiox xei^ievov xoivfjv Y^viav 
eXov auxo), Ttepl xf|v auxfjv 8id^exp6v eaxi xo3 oXo>. 

Ako yap KapaXXrjXoYpd^^iou xou ABrA TtapaXXrjXoYpa- 
(iuov dcp/jprjaOw xo AZ ojioiov iu ABrA xal 6^io(w<; 
xsi^ievov xoivrjv Y«viav £)(ov auxw xrjv utio AAB- Xey", 



K 




L E M 



Let ABC be the given rectilinear figure to which it is 
required to construct a similar (rectilinear figure), and D 
the (rectilinear figure) to which (the constructed figure) 
is required (to be) equal. So it is required to construct 
a single (rectilinear figure) similar to ABC, and equal to 
D. 

For let the parallelogram BE, equal to triangle ABC, 
have been applied to (the straight-line) BC [Prop. 1.44], 
and the parallelogram CM, equal to D, (have been ap- 
plied) to (the straight-line) CE, in the angle FCE, which 
is equal to CBL [Prop. 1.45]. Thus, BC is straight-on to 
CF, and LE to EM [Prop. 1.14]. And let the mean pro- 
portion GH have been taken of BC and CF [Prop. 6.13] . 
And let KGB, similar, and similarly laid out, to ABC 
have been described on GH [Prop. 6.18]. 

And since as BC is to GH, so GH (is) to CF, and if 
three straight-lines are proportional then as the first is to 
the third, so the figure (described) on the first (is) to the 
similar, and similarly described, (figure) on the second 
[Prop. 6.19 corn], thus as BC is to CF, so triangle ABC 
(is) to triangle KGH. But, also, as BC (is) to CF, so 
parallelogram BE (is) to parallelogram EF [Prop. 6.1]. 
And, thus, as triangle ABC (is) to triangle KGH, so par- 
allelogram BE (is) to parallelogram EF. Thus, alter- 
nately, as triangle ABC (is) to parallelogram BE, so tri- 
angle KGH (is) to parallelogram EF [Prop. 5.16]. And 
triangle ABC (is) equal to parallelogram BE. Thus, tri- 
angle KGH (is) also equal to parallelogram EF. But, 
parallelogram EF is equal to D. Thus, KGH is also equal 
to D. And KGH is also similar to ABC. 

Thus, a single (rectilinear figure) KGH has been con- 
structed (which is) similar to the given rectilinear figure 
ABC, and equal to a different given (rectilinear figure) 
D. (Which is) the very thing it was required to do. 

Proposition 26 

If from a parallelogram a(nother) parallelogram is 
subtracted (which is) similar, and similarly laid out, to 
the whole, having a common angle with it, then (the sub- 
tracted parallelogram) is about the same diagonal as the 
whole. 

For, from parallelogram ABCD, let (parallelogram) 



183 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



6xi uepl xrjv auxrjv Sidjiexpov eaxi to ABrA tw AZ. 



AH A 




b r 

Mr) yap, dXX' el 8uvax6v, eax« [auxfiv] 8id^iexpo<; fj 
AQr, xal expXrrdeTaa fj HZ SirjX'dw etxi xo 9, xal rj)cdco 
Sid xou oitopepa iwv AA, Br TtapdXXrjXoc; fj 0K. 

'Excel ouv uepl xrjv auxrjv Sidjuexpov eaxi xo ABrA iw 
KH, eaxiv dpa (be; rj A A npbz xrjv AB, ouxioc; f) HA upog 
xrjv AK. eaxi 8e xal 8ia xrjv o^toioxrjxa xwv ABrA, EH xal 
cb<; f) AA 7ip6<; xrjv AB, ouxck f] HA Tcp6<; xrjv AE- xal &>c, 
dpa f) HA npbc, xrjv AK, ouxcoc; fj HA npoc; xrjv AE. fj HA 
dpa TCpog exaxepav xfiv AK, AE xov auxov e/ei Xoyov. larj 
dpa eaxlv fj AE xfj AK fj eXdxxov xfj ^eii^ovi- orcep laxlv 
dBuvaxov. oux dpa oux eaxi uepl xrjv auxrjv 8id[uexpov xo 
ABrA xw AZ- uepl xrjv auxrjv dpa eaxi 8id^expov xo ABrA 
TiapaXXrjXoypa^ov xw AZ TtapaXXrjXoYpd^cp. 

'Edv dpa diio 7tapaXXrjXoYpd^[Uou TiapaXXrjXoypa^iuov 
dcpaipcdfj 6|ioi6v xe x« 6Xo xal b[io'ux>z xeijievov xoivrjv 
yoviav s^ov auxG, rcepl xrjv auxrjv Siduexpov eoxi x£3 6Xcp- 
oTiep e8ei 8eT^ai. 



ndvxwv xwv iiapd xrjv auxrjv eMeTav napapaXXo^ievwv 
TiapaXXrjXoYpd^i[Uwv xal eXXeinovxov eiSeai KapaXXrjXoypdfU- 
[ioic by.oioic, xe xal 6iuoio<; xeiiuevou; xw aito xfj? fjiuiaeiac; 
dvaypacpotuevw [ueyiaxov eaxi xo duo xff. rj^iiaeiac; napa- 
PaXXo^tevov [TtapaXXrjXoYpa^ov] o^ioiov ov iu eXXei^ifiavxi. 

"Eaxw eu'deTa fj AB xal xexurja'do "biya. xaxa xo T, 
xal TiapapepXrjadw Ttapa xrjv AB eu'delav xo AA uapaX- 
XrjXoYpaji^ov eXXelnov ei8ei TCapaXXrjXoypd^o xw AB dva- 
Ypacpevxi duo xrj<; rj[iiae(a<; xfj<; AB, xouxeaxi xrj<; TB - Xeya>, 
oxi udvxwv xwv Ttapa xrjv AB uapapaXXo^evwv TtapaXXrj- 
XoYpd^[Uwv xal eXXeiKovxwv eiSeai [TtapaXXrjXoypdfU^ou;] 
oiuoiok; xe xal 6uo(m<; xeiuevoic; xo AB ^icyiaxov eaxi xo 



AF have been subtracted (which is) similar, and similarly- 
laid out, to ABCD, having the common angle DAB with 
it. I say that ABCD is about the same diagonal as AF. 



AG D 




B C 

For (if) not, then, if possible, let ARC be [ABCD's] 
diagonal. And producing GF, let it have been drawn 
through to (point) H. And let HK have been drawn 
through (point) H, parallel to either of AD or BC 
[Prop. 1.31]. 

Therefore, since ABCD is about the same diagonal as 
KG, thus as DA is to AB, so GA (is) to AK [Prop. 6.24]. 
And, on account of the similarity of ABCD and EG, also, 
as DA (is) to AB, so GA (is) to AE. Thus, also, as GA 
(is) to AK, so GA (is) to AE. Thus, GA has the same 
ratio to each of AK and AE. Thus, AE is equal to AK 
[Prop. 5.9], the lesser to the greater. The very thing is 
impossible. Thus, ABCD is not not about the same di- 
agonal as AF. Thus, parallelogram ABCD is about the 
same diagonal as parallelogram AF. 

Thus, if from a parallelogram a(nother) parallelogram 
is subtracted (which is) similar, and similarly laid out, 
to the whole, having a common angle with it, then (the 
subtracted parallelogram) is about the same diagonal as 
the whole. (Which is) the very thing it was required to 
show. 

Proposition 27 

Of all the parallelograms applied to the same straight- 
line, and falling short by parallelogrammic figures similar, 
and similarly laid out, to the (parallelogram) described 
on half (the straight-line), the greatest is the [parallelo- 
gram] applied to half (the straight-line) which (is) similar 
to (that parallelogram) by which it falls short. 

Let AB be a straight-line, and let it have been cut in 
half at (point) C [Prop. 1.10]. And let the parallelogram 
AD have been applied to the straight-line AB, falling 
short by the parallelogrammic figure DB (which is) ap- 
plied to half of AB — that is to say, CB. I say that of all 
the parallelograms applied to AB, and falling short by 



184 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



AA. Tiocpotpepxyja'dco yap Tiapd xrjv AB euifteTav to AZ na- 
paXX/]X6ypa^ov eXXeinov d'Bsi TiapaXX/jXoypd^cp to ZB 
o^oto xe xal o^oiwc xei^tevw xcp AB- Xeyto, oxi uei£6v eaxi 
xo AA xou AZ. 



A E 




A r K B 

'Etc! yap ojioiov eaxi xo AB TtapaXXrjXoypa^ov xG ZB 
7iapaXX/]Xoypd[i^.CL), Ttepl x/]v auxr|v eiai 8id[iexpov. f))fdw 
auxwv 8id^.expo<; f] AB, xal xaxayeypdcp-dw xo ax?j^a. 

'Ercel o5v iaov eaxi xo TZ xw ZE, xoivov Be xo ZB, 
oXov dpa xo T6 oX« xw KE eaxiv iaov. dXXd xo T6 tu 
TH eaxiv iaov, etiei xal f] Ar xfj TB. xal xo BT dpa x£S EK 
eaxiv iaov. xoivov Ttpoaxe(a , da> xo TZ' oXov dpa xo AZ iu 
AMN yvw^tovi eaxiv iaov waxe xo AB TtapaXXr)X6ypa^iov, 
xouxeaxi xo AA, xou AZ 7iapaXX/]Xoypd[i^ou \±s%6v eaxiv. 

ndvxcov dpa xwv Ttapa xrjv auxrjv eu'delav TtapapaX- 
Xo^tevwv TiapaXXr)Xoypd^«v xal eXXeiTtovxov eiBeai TtapaX- 
Xr]Xoypd^oi<; b\±oioiq xe xal 6\±oig>z xei^iivou; iw duo xfj<; 
f|^iaeia<; dvaypacpo^tevo ^teyiaxov eaxi xo duo xrj<; f^iaeiac; 
7iapapX/]Tf)£v orcep e8ei BeT^ai. 



XT]'. 

Ilapd xiqv 8oi9eTaav euiMav xw Bo'devxi eu'duypd^w 
Iaov napaXXiqXoypa^ov napapaXelv eXXeutov ei8ei na- 
paXX/)Xoypd|jijicp 6|ioicp xw BoiSevxi- Bel Be xo BiBopievov 
eO'duypa^ov [S BeT iaov TtapapaXeTv] ^ir] ^xeT^ov eivai xou 
duo xrjg rpiaeiag dvaypacpo^ievou o^ioiou x« eXXei^axi [xou 
xe duo xfj? f)|jiiaeia<; xal 5 Bel o^ioiov eXXeineiv] . 

'Eaxco f) (iev So-deTaa eui)eTa f) AB, xo Be Bo-dev 
eu-duypajijiov, £> BeTTaov Ttapd xrjv AB TiapapaXelv, xo T [if] 
^el^ov [dv] xou dno xfjc; fj^iiaeiac; xfjg AB dvaypacpo^ievou 
o^tobu ifi eXXei^axi, S Be: BeT o^ioiov eXXeiTteiv, xo A- 8a Br) 



[parallelogrammic] figures similar, and similarly laid out, 
to DB, the greatest is AD. For let the parallelogram AF 
have been applied to the straight-line AB, falling short by 
the parallelogrammic figure FB (which is) similar, and 
similarly laid out, to DB. I say that AD is greater than 
AF. 



D E 




A C K B 

For since parallelogram DB is similar to parallelo- 
gram FB, they are about the same diagonal [Prop. 6.26]. 
Let their (common) diagonal DB have been drawn, and 
let the (rest of the) figure have been described. 

Therefore, since (complement) CF is equal to (com- 
plement) FE [Prop. 1.43], and (parallelogram) FB is 
common, the whole (parallelogram) CH is thus equal 
to the whole (parallelogram) KE. But, (parallelogram) 
CH is equal to CG, since AC (is) also (equal) to CB 
[Prop. 6.1]. Thus, (parallelogram) GC is also equal 
to EK. Let (parallelogram) CF have been added to 
both. Thus, the whole (parallelogram) AF is equal to 
the gnomon LMN. Hence, parallelogram DB — that is to 
say, AD — is greater than parallelogram AF. 

Thus, for all parallelograms applied to the same 
straight-line, and falling short by a parallelogrammic 
figure similar, and similarly laid out, to the (parallelo- 
gram) described on half (the straight-line), the greatest 
is the [parallelogram] applied to half (the straight-line). 
(Which is) the very thing it was required to show. 

Proposition 28* 

To apply a parallelogram, equal to a given rectilin- 
ear figure, to a given straight-line, (the applied parallel- 
ogram) falling short by a parallelogrammic figure similar 
to a given (parallelogram) . It is necessary for the given 
rectilinear figure [to which it is required to apply an equal 
(parallelogram)] not to be greater than the (parallelo- 
gram) described on half (of the straight-line) and similar 
to the deficit. 

Let AB be the given straight-line, and C the given 
rectilinear figure to which the (parallelogram) applied to 



185 



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ELEMENTS BOOK 6 



Ttapd xf]v BoiDeToav eMelav xfjv AB xG SoiSevxi eO'duypajji^cp 
xG F laov TtapaXXr)X6ypajj.(jiov TtapapaXeTv cXXcTttov d'Ssi rca- 
paXX/]Xoypd|jijKp o^ioup ovxi xG A. 



6 H O Z 




K 



Tex[if]OT}« f) AB 8[)(a xaxd xo E arjjieiov, xal dva- 
yeypacp'dco duo xfj? EB xG A o^toiov xal 6(ioiw<; xd^iervov 
xo EBZH, xal au^KETiXrjpwa'dw xo AH TtapaXXr)X6ypa^ov. 

EE \itv ouv iaov eaxl xo AH xG T, yeyovog av ei'r] xo era- 
xa)cdev 7tapapepXr]xai yap napa xf]v So-deTaav euiJeTav xf]v 
AB xG 8oi9evxi eui9uypd[i|jiw xG T laov TtapaXXrjXoypa^ov 
xo AH eXXeljtov eiBei 7tapaXX/]Xoypd|i^.tL> xG HB 6[io(cl> ovxi 
xG A. ei Se ou, [iziZov eaxto xo 0E xou T. 1'aov 8e xo 9E 
xG HB- [leiZov dpa xal xo HB xou T. G 8f] ^si£6v eaxi 
xo HB xou r, xauxr] xfj U7tepo)(fj 1'aov, xG 8e A ojioiov xal 
o^ioicx xei^ievov xo auxo auveaxdxo xo KAMN. dXXa xo A 
xG HB [eaxiv] o^toiov xal xo KM dpa xG HB eaxiv o^ioiov. 
eaxw ouv 6^i6Xoyo<; f] ^.ev KA xf) HE, f] 8e AM xfj HZ. 
xal snei laov eaxl xo HB xou; T, KM, ^.a^ov dpa eaxl xo 
HB xou KM- ^lei^wv dpa eaxl xal f) [lev HE xfj? KA, r] 8s 
HZ xfj? AM. xela-dw xrj ^tev KA Tar] f] HS, xfj 8e AM Xar] 
f] HO, xal aupi7i£7T;Xr]pGa , dw xo SHOn TtapaXXr)X6ypa^ov 
laov dpa xal o\±oiov eaxi [xo Hn] xG KM [dXXa xo KM xG 
HB 6\Loi6v eaxiv]. xal xo Hn dpa xG HB o^toiov eaxiv Ttepl 
xf]v auxrjv dpa Bid^texpov eaxi xo Hn xG HB. eax« auxGv 
Sid^texpoi; f] HnB, xal xaxayeypdcp^w xo a)(rjpc. 

'EtieI ouv 'laov eaxl xo BH xoI<; T, KM, Gv xo Hn xG 
KM eaxiv iaov, Xoitcoc dpa 6 TX$ yvo^iwv XoitcG xG T laoc, 
eaxiv. xal CTtel 'laov eaxl xo OP xG SE, xoivov npoaxeia'do 
xo HB' 6Xov dpa xo OB oXw xG SB i'aov eaxiv. dXXa xo SB 
xG TE eaxiv laov, etcei xal rcXeupa f] AE TtXeupa xfj EB eaxiv 
\ar\- xal xo TE dpa xG OB eaxiv i'aov. xoivov Ttpoaxeia'dw 
xo SE - 6Xov dpa xo TE oXw xG $XT yvG^tovi saxiv laov. 
dXX' 6 <3>XT yvG^twv xG V zbziyfti} I'aoc xal xo TE dpa xG 
T eaxiv iaov. 

napa xf)v SoiJelaav dpa euiDeTav xf]v AB xG Soif)evxi 
eu-duypd^jjiw xG V i'aov TiapaXX^Xoypa^ov TiapapepXrjxai 
xo ET eXXeliiov eiSei 7iapaXXr]Xoypd^w xG nB 6[ioiw ovxi 



AB is required (to be) equal, [being] not greater than 
the (parallelogram) described on half of AB and similar 
to the deficit, and D the (parallelogram) to which the 
deficit is required (to be) similar. So it is required to apply 
a parallelogram, equal to the given rectilinear figure C, to 
the straight-line AB, falling short by a parallelogrammic 
figure which is similar to D. 



H G P F 




K N 



Let AB have been cut in half at point E [Prop. 1.10], 
and let (parallelogram) EBFG, (which is) similar, and 
similarly laid out, to (parallelogram) D, have been de- 
scribed on EB [Prop. 6.18]. And let parallelogram AG 
have been completed. 

Therefore, if AG is equal to C then the thing pre- 
scribed has happened. For a parallelogram AG, equal 
to the given rectilinear figure C, has been applied to the 
given straight-line AB, falling short by a parallelogram- 
mic figure GB which is similar to D. And if not, let HE 
be greater than C. And HE (is) equal to GB [Prop. 6.1]. 
Thus, GB (is) also greater than C. So, let (parallelo- 
gram) KLMN have been constructed (so as to be) both 
similar, and similarly laid out, to D, and equal to the ex- 
cess by which GB is greater than C [Prop. 6.25]. But, 
GB [is] similar to D. Thus, KM is also similar to GB 
[Prop. 6.21]. Therefore, let KL correspond to GE, and 
LM to GF. And since (parallelogram) GB is equal to 
(figure) C and (parallelogram) KM, GB is thus greater 
than KM. Thus, GE is also greater than KL, and GF 
than LM. Let GO be made equal to KL, and GP to LM 
[Prop. 1.3]. And let the parallelogram OGPQ have been 
completed. Thus, [GQ] is equal and similar to KM [but, 
KM is similar to GB] . Thus, GQ is also similar to GB 
[Prop. 6.21]. Thus, GQ and GB are about the same diag- 
onal [Prop. 6.26]. Let GQB be their (common) diagonal, 
and let the (remainder of the) figure have been described. 

Therefore, since BG is equal to C and KM, of which 
GQ is equal to KM, the remaining gnomon UWV is thus 
equal to the remainder C. And since (the complement) 
PR is equal to (the complement) OS [Prop. 1.43], let 
(parallelogram) QB have been added to both. Thus, the 
whole (parallelogram) PB is equal to the whole (par- 



186 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



t<3 A [eTieiSyjTcep to IIB tw HII o^ioiov eaxiv]- oTtep eSei allelogram) OB. But, OB is equal to TT, since side 
noifjoai. AE is equal to side EB [Prop. 6.1]. Thus, TE is also 

equal to PB. Let (parallelogram) OS have been added 
to both. Thus, the whole (parallelogram) TS is equal to 
the gnomon VWU. But, gnomon VWU was shown (to 
be) equal to C. Therefore, (parallelogram) TS is also 
equal to (figure) C. 

Thus, the parallelogram ST, equal to the given rec- 
tilinear figure C, has been applied to the given straight- 
line AB, falling short by the parallelogrammic figure QB, 
which is similar to D [inasmuch as QB is similar to GQ 
[Prop. 6.24] ]. (Which is) the very thing it was required 
to do. 

t This proposition is a geometric solution of the quadratic equation x 2 — ax+/3 = 0. Here, x is the ratio of a side of the deficit to the corresponding 
side of figure D, a is the ratio of the length of AB to the length of that side of figure D which corresponds to the side of the deficit running along 
AB, and (3 is the ratio of the areas of figures C and D. The constraint corresponds to the condition < a 2 /A for the equation to have real roots. 
Only the smaller root of the equation is found. The larger root can be found by a similar method. 



ITocpa tt)V BcnMaav eru'deTocv tG BotJevti eufluYpd^cp 
'laov TTapaXXrjXoypajujiov TtapapaXsIv OrapPaXXov ri'Bsi ua- 
pocXXr)Xoypd^uw 6\ioi(x) xw Bo'devTi. 



A M 



r \ 

/ 

A E/ 


\ k / 

#Uo 


/ / 





e 



N 



n e h 

"Eaxw f] y.ev Bo'rMaa su-vMa f] AB, to 8e 8oi9sv 
eu-OuYpa^ov, CS 5ei I'aov rcapa tt)v AB TtapapaXsIv, to T, 
S 8e Bel 6[ioiov UTteppdXXeiv, to A' 8eT 8f) rcapa tt]v AB 
sMeiav tw T eu^UYpd^iico laov 7i:apaXXr]X6Ypa|jijj.ov napa- 
PaXsiv UTieppdXXov ri'Bsi 7tapaXX/]XoYpd|ji^to ofioicp tc5 A. 

TeT[i.r)ai[)co f) AB B[)(a xaTa to E, xod dvayeYpd'&o 
duo t/)c EB to A ojioiov xal ojioimc; xri^iervov rcapaX- 
X/]XoYpa^tiov to BZ, xal auvapicpoTepoig ^tev toT<; BZ, T 
laov, to 8e A 6[ioiov xdi 6[ioia>c xsifisvov to ocuto au- 
vEOTaTW to H6. o^ioXoyoc; 8e eaTCO f) ^ev K6 Tfj ZA, f) 8e 
KH Tfj ZE. xdi ETte! ^£i£6v ecru to H0 toO ZB, ^d^wv dpa 
ecttI xdi f] ^lev K9 Tfj? ZA, f] 8s: KH Tfj ZE. expepXf]a , doaav 
di ZA, ZE, xdi Tfj [iev K0 for) soto f) ZAM, Tfj 8e KH for) 
f] ZEN, xdi au^7i£TtXr)pcfo , do to MN- to MN dpa iw H9 
laov ts eaTi xal ojioiov. dXXa to H0 t£5 EA sgtiv o^oiov 



Proposition 29 f 

To apply a parallelogram, equal to a given rectilin- 
ear figure, to a given straight-line, (the applied parallelo- 
gram) overshooting by a parallelogrammic figure similar 
to a given (parallelogram) . 



L M 



H 



\ c 













A 


El 


x\fr, L 


/ / IN 



N Q O G 

Let AB be the given straight-line, and C the given 
rectilinear figure to which the (parallelogram) applied to 
AB is required (to be) equal, and D the (parallelogram) 
to which the excess is required (to be) similar. So it is 
required to apply a parallelogram, equal to the given rec- 
tilinear figure C, to the given straight-line AB, overshoot- 
ing by a parallelogrammic figure similar to D. 

Let AB have been cut in half at (point) E [Prop. 1.10], 
and let the parallelogram BF, (which is) similar, and 
similarly laid out, to D, have been described on EB 
[Prop. 6.18]. And let (parallelogram) GH have been con- 
structed (so as to be) both similar, and similarly laid out, 
to D, and equal to the sum of BF and C [Prop. 6.25]. 
And let KH correspond to FL, and KG to FE. And since 
(parallelogram) GH is greater than (parallelogram) FB, 



187 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



xal to MN dpa tG EA 6[ioi6v egtiv Ttepl ttjv auxrjv dpa 
8id|jieTp6v eoxi to EA to MN. f^x'dw auTGv Bid^iETpog #) 
ZS, xai xaTayeYP&P'fl" TO a X^ a - 

'EtteI Xaov sotI to H6 toTc; EA, T, dXXd to H6 tG MN 
iaov eotiv, xal to MN dpa toTc; EA, T Xaov eotlv. xoivov 
dcprjprp'dw to EA- Xoittoc; dpa 6 ^/X<i> Y^Gpiwv tG T sotiv 
I'oog. xal etieI iar) egtIv f) AE xfj EB, iaov eoti xal to AN tG 
NB, touteotl tG AO. xoivov TtpoaxEiadw to EE 1 oXov dpa 
to AS iaov eot ! tG yvGjiovi. dXXd 6 f&X^ yvG^icov 

tG r "taoc, eaTiv xal to AE dpa tG T Xaov eotiv. 

Ilapa Trjv ScdeTaav dpa sui&ETav tt]v AB tG BotSevti 
eru'duypafji^cp tG T Xaov KapaXXrjXoYpotji^ov TtapapspXrjTai 
to AS ujieppdXXov eiSei TCapaXXrjXoYpdjijKp tG nO ojioicp 
ovti tG A, etisI xal tG EA sotiv ojioiov to On- orap e6el 
Tioifjaai. 



KH is thus also greater than FL, and KG than FE. 
Let FL and FE 1 have been produced, and let FLM be 
(made) equal to KH, and FEN to ifG [Prop. 1.3]. And 
let (parallelogram) MN have been completed. Thus, 
MN is equal and similar to GH. But, GH is similar to 
EL. Thus, MJV is also similar to EL [Prop. 6.21]. EL is 
thus about the same diagonal as MN [Prop. 6.26]. Let 
their (common) diagonal FO have been drawn, and let 
the (remainder of the) figure have been described. 

And since (parallelogram) GH is equal to (parallel- 
ogram) EL and (figure) C, but GH is equal to (paral- 
lelogram) MN, MN is thus also equal to EL and C. 
Let EL have been subtracted from both. Thus, the re- 
maining gnomon XWV is equal to (figure) C. And since 
AE is equal to EB, (parallelogram) AN is also equal to 
(parallelogram) NB [Prop. 6.1], that is to say, (parallel- 
ogram) LP [Prop. 1.43]. Let (parallelogram) EO have 
been added to both. Thus, the whole (parallelogram) AO 
is equal to the gnomon VWX. But, the gnomon VWX 
is equal to (figure) C. Thus, (parallelogram) AO is also 
equal to (figure) C. 

Thus, the parallelogram AO, equal to the given rec- 
tilinear figure C, has been applied to the given straight- 
line AB, overshooting by the parallelogrammic figure QP 
which is similar to D, since PQ is also similar to EL 
[Prop. 6.24] . (Which is) the very thing it was required 
to do. 



t This proposition is a geometric solution of the quadratic equation x 2 +a x — (3 = 0. Here, x is the ratio of a side of the excess to the corresponding 
side of figure D, a is the ratio of the length of AB to the length of that side of figure D which corresponds to the side of the excess running along 
AB, and (3 is the ratio of the areas of figures C and D. Only the positive root of the equation is found. 



X' 



Proposition 30+ 



Trjv SoiMoav sudslav TteTCpaa|jievr]v dxpov xal (isaov To cut a given finite straight-line in extreme and mean 
Xoyov te^eTv. ratio. 



r 



z o 



H 



A 



J B 



J B 



'A 

TEotw f) So-dsTaa EU-dsTa K£7i£paa[i£vr) f] AB- 5eT 8f] ttjv 
AB EtrdsTav dxpov xal ^saov Xoyov te^eiv. 



D 

Let AB be the given finite straight-line. So it is re- 
quired to cut the straight-line AB in extreme and mean 



188 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



AvayeypdcpiDw duo xfj<; AB xexpdyiovov to Br, xal rca- 
paPepXr]o , dw napd xr)v Ar xco Br iaov 7iapaXXr]X6ypa[j.(jiov 
xo TA (meppdXXov el'8ei xw AA o^xoicp iw Br. 

Texpdycovov 8e eaxi xo Br- xexpdywvov dpa eaxi xdi 
xo AA. xdi £7td laov eaxi xo Br xw TA, xoivov dcp/]pr)aiL>M 
xo TE- Xomov dpa xo BZ XoiraS xcp AA eaxiv '(aov. eaxi 
8s auxw xdi laoywviov xfiv BZ, AA dpa avxiTiCKovdaaiv ai 
TiXeupal ai nepl xd<; iaa<; ytoviac eaxiv dpa ci><; f) ZE -rcpoc; 
xr]v EA, ouxwc; f) AE npo? xrjv EB. Tar) Se f] ^iev ZE xrj AB, 
f] 8e EA xfj AE. eaxiv dpa &>c, rj BA rcpoc; x/]v AE, ouxgx f) 
AE Ttpoc; xfjv EB. ^iei£«v 5e f] AB xrjc; AE- ^ei^tov dpa xal 
r) AE xfjc; EB. 

H dpa AB euiDeTa dxpov xal ^.eaov Xoyov xex^irjxai xaxd 
xo E, xal xo ^.elCov auxrj<; x[ifj^.d eaxi xo AE- oitep e8ei 
Ttoirjaai. 



ratio. 

Let the square BC have been described on AB [Prop. 
1.46], and let the parallelogram CD, equal to BC, have 
been applied to AC, overshooting by the figure AD 
(which is) similar to BC [Prop. 6.29]. 

And BC is a square. Thus, AD is also a square. 
And since BC is equal to CD, let (rectangle) CE have 
been subtracted from both. Thus, the remaining (rect- 
angle) BF is equal to the remaining (square) AD. And 
it is also equiangular to it. Thus, the sides of BF and 
AD about the equal angles are reciprocally proportional 
[Prop. 6.14]. Thus, as FE is to ED, so AE (is) to EB. 
And FE (is) equal to AB, and ED to AE. Thus, as BA is 
to AE, so AE (is) to EB. And AB (is) greater than AE. 
Thus, AE (is) also greater than EB [Prop. 5.14]. 

Thus, the straight-line AB has been cut in extreme 
and mean ratio at E, and AE is its greater piece. (Which 
is) the very thing it was required to do. 



t This method of cutting a straight-line is sometimes called the "Golden Section" — see Prop. 2.11. 



Xa'. 

'Ev xou; opiJoycovioic; xpiyiovoic; xo duo xfjc; xr)v op'drjv 
ywviav U7ioxeivouar]<; TtXeupac; eI8o<; I'aov eaxi xou; duo xwv 
xr]v 6pi}r]v ywviav Ttepiexouawv TtXeupov ei5eai xolg ojioioic; 
xe xal o^oicoc; dvaypacpouevoic;. 




'Eaxw xpiywvov op-doywviov xo ABr op'drjv e)(ov xrjv 
Otio BAT yoviav Xeyw, oxi xo anb xfjc BT eTSoc 'laov eaxi 
xolc; duo xfiv BA, Ar eToeai xoI<; ouoiok; xe xal 6uo(w<; 
dvaypacpo^ievoic;. 

"H)(Tf)w xd'dexoc; f) AA. 

'Etcei oov ev opiDoywviw xpiywvw ifi ABr anb xrjc npbc, 
xcp A opi^c; yoviac; era xrjv Br pdaiv xd-fJexoc; rjxxai f\ AA, 
xd ABA, AAr Ttpo? xfj xaiSexcp xpiywva opioid eaxi x£> xe 
0X0 xw ABr xal dXXr)Xou;. xal enel o^ioiov eaxi xo ABr xS 
ABA, eaxiv dpa cb<; f) TB 7tp6<; xfjv BA, ouxwc; f) AB npo? 
xr]v BA. xal CTiel xpeTc eui9eTai dvdXoyov eiaiv, eaxiv cbc; f) 
npd>xr) 7ip6<; xrjv xpixrjv, ouxw<; xo duo xrjc; 7tpcoxr]C eTBoc Tipoc 



Proposition 31 

In right-angled triangles, the figure (drawn) on the 
side subtending the right-angle is equal to the (sum of 
the) similar, and similarly described, figures on the sides 
surrounding the right-angle. 




Let ABC be a right-angled triangle having the angle 
BAC a right-angle. I say that the figure (drawn) on BC is 
equal to the (sum of the) similar, and similarly described, 
figures on BA and AC. 

Let the perpendicular AD have been drawn [Prop. 
1.12]. 

Therefore, since, in the right-angled triangle ABC, 
the (straight-line) AD has been drawn from the right- 
angle at A perpendicular to the base BC, the trian- 
gles ABD and ADC about the perpendicular are sim- 
ilar to the whole (triangle) ABC, and to one another 
[Prop. 6.8]. And since ABC is similar to ABD, thus 



189 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



to duo xfjg Bsuxspag to ojioiov xal o^ioicoc; dvaypacpojisvov. 
(be; dpa f) TB rcpoc; xrjv BA, ouxwc; xo duo xfjc; TB eT8oc; 
Tipog xo duo xfjc BA xo ojioiov xdi ojioiwc; dvaypacpo^iEvov. 
8ia xd auxd 6f) xdi (be; f) Br Tipoc; xrjv TA, ouxcog xo drco xrj<; 
Br sTBoc; Ttpoc; xo duo xfjc TA. <baxe xdi (be; f) Br upog xdg 
BA, Ar, oux(oc; xo arco xrjg Br eTSoc; upoc; xd duo x£5v BA, 
Ar xd ojjioia xdi ojioitog dvaypacpojieva. Tar] Bs rj Br xdu; 
BA, Ar- Taov dpa xdi xo duo xfjc Br eiSog xou; duo xwv 
BA, Ar elbeai xou; ojioioic; xs xdi o^iolioc; dvaypacpo^ievoic;. 

'Ev dpa xolg op'doywvioic; xpiywvou; xo a.nb xfjc xrjv 
opiSrjv ywviav unoxeivouarjc; TtXcupdc; eT6oc; Taov eaxl xou; 
dTio xwv xrjv opi^rjv yioviav TiepiexouaSSv nXeupfiv ri'Bsai xoTg 
o^oiou; xs xal o^oiwg dvaypacpojis vou;- onep eBei 8eic;ai. 



X(3'. 

'Edv 80o xpiyova auvxrdfj xaxd (iiav ywviav xdg 860 
TtXeupdc; xdu; 8ual TtXeupau; dvdXoyov £)(ovxa (baxs xdg 
opioXoyouc; otuxwv TtXsupdc xal 7iapaXXr]Xou<; elvai, ai Xomal 
xwv xpiywvwv TiXeupal z% sundae; eaovxai. 



A 




Br e 

'Eaxaj 860 xpiywva xd ABr, ArE xd? 860 rcXeupdc xdc 
BA, Ar xdu; 8ual rcXeupau; xdu; Ar, AE dvdXoyov s/ovxa, 
(be ^iev xrjv AB Tip6<; xrjv Ar, oux«<; xrjv Ar Ttpoc; xrjv AE, 
TiapdXXrjXov 8s xrjv (iev AB xrj Ar, xrjv 8e Ar xfj AE- Xsyco, 
oxi £7t' sO'dslac saxlv f] Br xfj TE. 

'Etce! yap napdXXrjXoc; eaxiv fj AB xrj Ar, xal tic, auxd? 
ejUTtSTtxwxev eu'deTa fj AT, ai evaXXd<; ywviai ai bnb BAT, 
ATA i'aai dXXrjXaic; riaiv. 81a xd auxd 8fj xal fj \mb TAE xfj 
O716 ABA iarj saxiv. (benx xal fj 6716 BAr xrj utco TAE iaxiv 
larj. xal ETiel 860 xpiywvd eaxi xd ABr, ArE puav ywviav 
xfjv Tipoc; xw A [iia ywvia xfj rcpoc xo A larjv s^ovxa, Tiepl 



as CB is to BA, so ,45 (is) to BD [Def. 6.1]. And 
since three straight-lines are proportional, as the first is 
to the third, so the figure (drawn) on the first is to the 
similar, and similarly described, (figure) on the second 
[Prop. 6.19 corr.]. Thus, as CB (is) to BD, so the fig- 
ure (drawn) on CB (is) to the similar, and similarly de- 
scribed, (figure) on BA. And so, for the same (reasons), 
as BC (is) to CD, so the figure (drawn) on BC (is) to 
the (figure) on CA. Hence, also, as BC (is) to BD and 
DC, so the figure (drawn) on BC (is) to the (sum of the) 
similar, and similarly described, (figures) on BA and AC 
[Prop. 5.24]. And BC is equal to BD and DC. Thus, the 
figure (drawn) on BC (is) also equal to the (sum of the) 
similar, and similarly described, figures on BA and AC 
[Prop. 5.9]. 

Thus, in right-angled triangles, the figure (drawn) on 
the side subtending the right-angle is equal to the (sum of 
the) similar, and similarly described, figures on the sides 
surrounding the right-angle. (Which is) the very thing it 
was required to show. 

Proposition 32 

If two triangles, having two sides proportional to two 
sides, are placed together at a single angle such that the 
corresponding sides are also parallel, then the remaining 
sides of the triangles will be straight-on (with respect to 
one another). 



D 




B C E 

Let ABC and DCE be two triangles having the two 
sides BA and AC proportional to the two sides DC and 
DE—so that as AB (is) to AC, so DC (is) to DE— and 
(having side) AB parallel to DC, and AC to DE. I say 
that (side) BC is straight-on to CE. 

For since AB is parallel to DC, and the straight-line 
AC has fallen across them, the alternate angles BAC and 
ACD are equal to one another [Prop. 1.29]. So, for the 
same (reasons), CDE is also equal to ACD. And, hence, 
BAC is equal to CDE. And since ABC and DCE are 
two triangles having the one angle at A equal to the one 



190 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



8s tolc, XaoLc, ycoviac; Tag TtXeupac; dvdXoyov, toe; xr)V BA Ttpoc; 
ttjv Ar, outoc; ttjv TA Ttpoc; xr)v AE, iaoycjviov apa eoxl 
to ABr Tpiycovov tw ArE Tpiywvw- for) apa f] utio ABr 
ywvia Tfj utio ArE. e8eix'dr) 8s xai f] utio ArA Tfj utco BAr 
for)- oXr] apa f] utio ArE 8uai TaT? utio ABr, BAr for) ecrciv. 
xoivf] Tipoaxeicydto f] utio ATB- ai apa utio ArE, ArB toic 
utio BAr, ArB, TBA foai slaw. dXX" ai utio BAr, ABr, 
ArB Suoiv op'ddic; foai eioiv xal ai utio ArE, ArB apa 
8uaiv op'ddic; foai siaiv. Tipoc; 8f] tivi cu-deia Tfj Ar xai iw 
Tcp6<; auTfj ar^eiw iu T 8uo eu-delai ai Br, TE [irj etc! t& 
auTa \iepr) xei^tevai iac, ecpec;rj<; ™c, utco ArE, ArB 

8uaiv op'ddic; foac; TioioOaiv etc' eu'deiac; apa ecrciv f] Br Tfj 

rE. 

°Edv apa 8uo Tpiycova auvTcdfj xaTa ^.iav ywviav Tag 
8uo TiXeupdc Talc; Suai TtXeupau; dvdXoyov exovTa wots tolc, 
o^ioXoyouc; auTWv TtXeupac; xai TiapaXXr)Xou<; elvai, ai XoiTial 
twv Tpiy«v«v TtXeupai etc' eu'deiac; eaovTai- oTtep eBei BeT^ai. 



Xy'. 

'Ev toT<; Took; xuxXoic; ai ywviai tov aikov e)(0UGi 
Xoyov tocT<; Ttepicpepeiaic;, ecp'' Sv peprjxaaiv, edv tc Tipoc; 
tou; xevTpoic; edv tc Ttpoc; tolc Ttepicpepeiaic; Sai pepr)xuiai. 




'Ecrccoaav fooi xuxXoi oi ABr, AEZ, xal Ttpoc; ^ev to!<; 
xevTpoic auTGv toTc; H, ywviai eoiuoav ai utco BBT, 
E6Z, Ttpoc 5e Talc; Tiepicpepeiau; ai utio BAr, EAZ- Xeyco, 
oti ecrciv cbc; f) BV Ttepicpepeia Ttpoc; t/]v EZ Tiepicpepeiav, 
outox fj tc utio BHT yovia Ttpoc; t/)v utio E6Z xai f) utco 
BAr Ttpoc; tt)v utio EAZ. 

Keicydtoaav yap Tfj [ie\i BT Ttepicpepeia foai xaTa to ec;rjc; 
6aai8r)TtoTouv ai TK, KA, Tfj 8e EZ Tiepicpepeia foai oaai- 
8r)TtoTouv ai ZM, MN, xai CTteCeux'dwaav ai HK, HA, 0M, 
6N. 

'Etcei ouv i'aai eiaiv ai Br, TK, KA Tiepicpepeiai dXXf|Xaic;, 
foai eiai xai ai utio BBT, THK, KHA ywviai dXXrjXaic 
oaaTtXaaiov apa ecrciv f] BA Ttepicpepeia xfj? Br, ToaauTa- 
TtXaaiwv ecrci xai f] utio BHA yuvia Tfj? utio BHr. 8id Ta 



angle at D, and the sides about the equal angles pro- 
portional, (so that) as BA (is) to AC, so CD (is) to 
DE, triangle ABC is thus equiangular to triangle DCE 
[Prop. 6.6]. Thus, angle ABC is equal to DCE. And (an- 
gle) ACD was also shown (to be) equal to BAC. Thus, 
the whole (angle) ACE is equal to the two (angles) ABC 
and BAC. Let ACB have been added to both. Thus, 
ACE and ACB are equal to BAC, ACB, and CBA. 
But, BAC, ABC, and ACB are equal to two right-angles 
[Prop. 1.32]. Thus, ACE and ACB are also equal to two 
right-angles. Thus, the two straight-lines BC and CE, 
not lying on the same side, make adjacent angles ACE 
and ACB (whose sum is) equal to two right-angles with 
some straight-line AC, at the point C on it. Thus, BC is 
straight-on to CE [Prop. 1.14]. 

Thus, if two triangles, having two sides proportional 
to two sides, are placed together at a single angle such 
that the corresponding sides are also parallel, then the 
remaining sides of the triangles will be straight-on (with 
respect to one another). (Which is) the very thing it was 
required to show 

Proposition 33 

In equal circles, angles have the same ratio as the (ra- 
tio of the) circumferences on which they stand, whether 
they are standing at the centers (of the circles) or at the 
circumferences. 




Let ABC and DEF be equal circles, and let BGC and 
EHF be angles at their centers, G and H (respectively), 
and BAC and EDF (angles) at their circumferences. I 
say that as circumference BC is to circumference EF, so 
angle BGC (is) to EHF, and (angle) BAC to EDF. 

For let any number whatsoever of consecutive (cir- 
cumferences), CK and KL, be made equal to circumfer- 
ence BC, and any number whatsoever, FM and MN, to 
circumference EF. And let GK, GL, HM, and HN have 
been joined. 

Therefore, since circumferences BC, CK, and KL are 
equal to one another, angles BGC, CGK, and KGL are 
also equal to one another [Prop. 3.27]. Thus, as many 
times as circumference BL is (divisible) by BC, so many 



191 



ETOIXEIfiN 9'. 



ELEMENTS BOOK 6 



auxd 8f] xal oaaTcXaatav eaxlv rj NE Tcepicpepeia xfjc EZ, xo- 
aauxaTcXaalov eaxl xal f] utco N0E ytovia xfj? utco E0Z. si 
apa for) eaxlv f] BA Tcepicpepeia xfj EN Tcepicpepeia, !ar] eaxl 
xal ycovia f\ utco BHA xfj utio EON, xal el [iei^tov eaxlv f] BA 
Tiepicpepeia xfj? EN Tcepicpepeia?, jiei^Mv eaxl xal f) utio BHA 
ycovla xfj? utio EON, xal el eXdaawv, eXdaawv. xeaadpwv 
8f| ovxov jieye-dfiiv, Suo (lev itepicpepeiwv xfiv Br, EZ, 660 
8e ywvifiv x£Sv utio BBT, E6Z, eTXr)Tixai xfj? ^iev Br Tiepi- 
cpepeia? xal xfjt; utco BHr ycovia? ladxi? TcoXXaTcXaalwv fj xe 
BA Tcepicpepeia xal r\ utco BHA y«via, xfj? 8e EZ Tcepicpepeia? 
xal xfj? utco E0Z ywvia? fj xe EN Tcepicpepia xal f\ utco E9N 
ycovia. xal 8e8eixxai, 6x1 el UTcepe^ei f\ BA Tcepicpepeia xfj? 
EN Tcepicpepeia?, UTcepexei xal rj utco BHA ytovla xfj? utco 
EON ywvla?, xal el Tar), Tar), xal el eXdaawv, eXdaawv. 
eaxiv apa, cb? f] Br Tcepicpepeia Tcpo? xf]v EZ, ouxco? f] utco 
BHr ywvia Tcpo? xf]V utco EOZ. dXX' <i>? f] utco BHr ywvia 
Tcpo? xfjv utco EOZ, ouxo? f\ utco BAr Tcpo? xf)V utco EAZ. 
8iTcXaa(a yap exaxepa exaxepa?. xal 6? apa f] Br Tcepicpepeia 
Tcpo? xf]V EZ Tcepicpepeiav, ouxw? fj xe utco BHr yovia Tcpo? 
xf]v utco E0Z xal f] utco BAr Tcpo? xf)v utco EAZ. 

'Ev apa xol? Taoi? xuxXoi? ai ywviai xov auxov exouai 
Xoyov xal? Tcepicpepelai?, ecp' Sv pepfjxaaiv, edv xe Tcpo? xoT? 
xevxpoi? edv xe Tcpo? xal? Tcepicpepelai? c5ai peprjxuTai - oTcep 
e8ei 8eT<;ai. 



t This is a straight-forward generalization of Prop. 3.27 



times is angle BGL also (divisible) by BGC. And so, for 
the same (reasons), as many times as circumference NE 
is (divisible) by EF, so many times is angle NHE also 
(divisible) by EHF. Thus, if circumference BL is equal 
to circumference EN then angle BGL is also equal to 
EHN [Prop. 3.27], and if circumference BL is greater 
than circumference EN then angle BGL is also greater 
than EHN J and if (BL is) less (than EN then BGL is 
also) less (than EHN). So there are four magnitudes, 
two circumferences BC and EF, and two angles BGC 
and EHF. And equal multiples have been taken of cir- 
cumference BC and angle BGC, (namely) circumference 
BL and angle BGL, and of circumference EF and an- 
gle EHF, (namely) circumference EN and angle EHN. 
And it has been shown that if circumference BL exceeds 
circumference EN then angle BGL also exceeds angle 
EHN, and if (BL is) equal (to EN then BGL is also) 
equal (to EHN), and if (BL is) less (than EN then BGL 
is also) less (than EHN). Thus, as circumference BC 
(is) to EF, so angle BGG (is) to EHF [Dei. 5.5]. But as 
angle BGC (is) to EHF, so (angle) BAG (is) to EDF 
[Prop. 5.15]. For the former (are) double the latter (re- 
spectively) [Prop. 3.20]. Thus, also, as circumference BC 
(is) to circumference EF, so angle BGC (is) to EHF, 
and BAG to EDF. 

Thus, in equal circles, angles have the same ratio as 
the (ratio of the) circumferences on which they stand, 
whether they are standing at the centers (of the circles) 
or at the circumferences. (Which is) the very thing it was 
required to show. 



192 



ELEMENTS BOOK 7 



Elementary Number Theory^ 



tThe propositions contained in Books 7-9 are generally attributed to the school of Pythagoras. 



193 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



"Opoi. 

a'. Movdc; saxiv, xai9' fjv sxaaxov x65v ovxcov §v XiyeTw.. 
P'. Api/d^ioc; 8s to ex jiovd8«v auyxei|jievov TiXfj'dot;. 
y'. Mepog eaxlv dpi/d^ioc; dpi/d^ioO 6 eXdaawv xou 
(lei^ovot;, oxav xaxajiexpfj xov [isic^ova. 
8'. Mspr) 5s, oxav y.r\ xaxajisxp/j. 

z. noXXaTiXdaiog Ss 6 [isi^cov xou sXdaaovoc;, oxav xa- 
xajjiexpfjxai Otto xou eXdaaovoc;. 

'Apxioc; dpi%6c; soxlv 6 Si/a Siaipoujievog. 

C. Ilepiaaoc; 8e 6 fjirj Biaipou^evot; Si^a f\ [6] jiovdBi 
Siacpspwv dpxiou dpidjjioO. 

/)'. Apxidxic; apxioc; dpi%6c; eaxiv 6 utto dpxbu dpidjiou 
jiexpoujievo^ xaxd dpxiov dpidjiov. 

'Apxidxic; 8e Ttepiaaoc; eaxiv 6 uuo dpxiou dpid[iou 
jjiexpoujievog xaxd itspiaaov dpiiSjiov. 

i'. Ilspiaadxic; 8s Ttepiaaoc; dpi%6c; eaxiv 6 Otco Ttepiaaou 
dpiiSpiou ^expou^ievoc; xaxd nepiaaov dpid^iov. 

ia'. IlpGxoc; dpid^ioc; eaxiv 6 [iovd8i \i6vx] ^.expou^tevoc;. 

ip'. ilpwxoi icpoc; dXXf]Xou<; dpid^oi eiaiv oi ^tovdSi [i6vr] 
^lexpou^tevoi xoivG ^texpw. 

iy'. Suvdexoc; dpi'd^oc; eaxiv 6 apidjiw xivi fjiexpou^ievoc;. 

18'. £uv$exoi 8e upoc; dXXf]Xou<; dpi-dnoi eiaiv oi dpi%« 
xivi ^lexpou^ievoi xoivfi ^.expw. 

ie'. Aprdjioc; dprd^iov TcoXXanXaaid^eiv XcYexai, oxav, 
oaai elolv ev auxw piovd8ec;, xoaauxdxic; auvxei}/] 6 ttoX- 
XajcXaaia^o^ievoc;, xal xe\ir\Ta.i xu;. 

if'. "Oxav 8e Suo dpi'&[Jiol TtoXXajcXaaidaavxec; dXXr|Xouc; 
jcoiwai xiva, 6 yevojjlevo^ eraTceBoc; xaXeTxai, nXeupal Be: 
auxou oi TcoXXaiiXaaidaavxec; dXXr|Xouc; dpid^ioi. 

iC- "Oxav 8e xpeTc; dpid^iol TcoXXaiiXaaidaavxec; dXXr|Xouc; 
Ttoifiai xiva, 6 Yevo^evog axepeoc; eaxiv, nXeupal Se auxou 
oi TroXXaTtXaaidaavxec; dXXr|Xouc; dpidjioL 

irj'. TexpdY«vog dpii9[i6c eaxiv 6 iadxic; i'aoc; f\ [6] uno 
Suo Taiov dpidjifiv Ttepie/oijievoc;. 

it}'. Kupog 8s 6 iadxic; laog iadxic; fj [6] utto xpifiv lacov 
dpii9|jL«v Tiepie)(6^evoc;. 

x'. Apiduoi dvdXoYov eiaiv, oxav 6 Ttpwxoc; xou 8euxepou 
xal 6 xpixoc; xou xexdpxou iadxic; rj TtoXXaTiXdaioc; fj xo auxo 
(icpoc; rj xa auxd fiepr) waiv. 

xa'. "O[ioioi eixLTxeBoL xai axepeoi dpid^ioi eiaiv oi 
avdXoYov £)(ovx£<; xa<; TiXeupdt;. 

xp'. TeXsioc dpi-djiot; eaxiv 6 xolz eauxou (iepeaiv laoc, 



Definitions 

1. A unit is (that) according to which each existing 
(thing) is said (to be) one. 

2. And a number (is) a multitude composed of units. t 

3. A number is part of a(nother) number, the lesser of 
the greater, when it measures the greater. * 

4. But (the lesser is) parts (of the greater) when it 
does not measure it. § 

5. And the greater (number is) a multiple of the lesser 
when it is measured by the lesser. 

6. An even number is one (which can be) divided in 
half. 

7. And an odd number is one (which can) not (be) 
divided in half, or which differs from an even number by 
a unit. 

8. An even-times-even number is one (which is) mea- 
sured by an even number according to an even number. ^ 

9. And an even-times-odd number is one (which 
is) measured by an even number according to an odd 
number.* 

10. And an odd-times-odd number is one (which 
is) measured by an odd number according to an odd 
number. 8 

11. A primed number is one (which is) measured by a 
unit alone. 

12. Numbers prime to one another are those (which 
are) measured by a unit alone as a common measure. 

13. A composite number is one (which is) measured 
by some number. 

14. And numbers composite to one another are those 
(which are) measured by some number as a common 
measure. 

15. A number is said to multiply a(nother) number 
when the (number being) multiplied is added (to itself) 
as many times as there are units in the former (number), 
and (thereby) some (other number) is produced. 

16. And when two numbers multiplying one another 
make some (other number) then the (number so) cre- 
ated is called plane, and its sides (are) the numbers which 
multiply one another. 

17. And when three numbers multiplying one another 
make some (other number) then the (number so) created 
is (called) solid, and its sides (are) the numbers which 
multiply one another. 

18. A square number is an equal times an equal, or (a 
plane number) contained by two equal numbers. 

19. And a cube (number) is an equal times an equal 
times an equal, or (a solid number) contained by three 
equal numbers. 



194 



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20. Numbers are proportional when the first is the 
same multiple, or the same part, or the same parts, of 
the second that the third (is) of the fourth. 

21. Similar plane and solid numbers are those having 
proportional sides. 

22. A perfect number is that which is equal to its own 
parts 



tt 



t In other words, a "number" is a positive integer greater than unity. 

t In other words, a number a is part of another number b if there exists some number n such that na = b. 

§ In other words, a number a is parts of another number b (where a < b) if there exist distinct numbers, m and n, such that na = mb. 

' In other words, an even-times-even number is the product of two even numbers. 

* In other words, an even-times-odd number is the product of an even and an odd number. 

$ In other words, an odd-times-odd number is the product of two odd numbers. 

II Literally, "first". 

tt In other words, a perfect number is equal to the sum of its own factors. 



Auo dpid^cov dviatov exxeijievwv, dvducpaipoujisvou 8s 
dsl xou sXdaaovoc; oltzo xou [idCovoc;, sav 6 Xsmo^isvoc; 
^r)Ss7toxs xaxajisxpfj xov npo sauxou, scoc oo Aeicp-drj jiovdc;, 
ol eZ, dp/rjc; dpi%ol TtpGxoi Ttpoc; dXXrjXouc; saovxai. 



At 



Proposition 1 

Two unequal numbers (being) laid down, and the 
lesser being continually subtracted, in turn, from the 
greater, if the remainder never measures the (number) 
preceding it, until a unit remains, then the original num- 
bers will be prime to one another. 



B 1 



r 

H 



F - 



H 



tC 
G 



Auo yap [dv(o«v] dpidjjicov xfiv AB, TA dvducpai- 
poujisvou del xou eXdaaovoc; dno xou jisi^ovoc; 6 Xsittojjievoc; 
(jirjBsTtoxs xaxa^sxpsixc.) xov npo sauxou, sex; ou Xsicp-dfj 
(iovdc Xeyw, oxi ol AB, TA TtpGxoi Ttpoc; dXXf]Xou<; slaiv, 
xouxsaxiv oxi xouc; AB, TA \xovclq \io\>r\ (jisxpsl. 

El ydp [ir\ siaiv ol AB, TA npfixoi npoc; dXXrjXouc;, 
^sxpiqasi tic; auxouc; dpi'd^oc;. [isxpeixw, xdi eaxw 6 E- xal 6 
[isv TA xov BZ ^isxpwv Xeittetio sauTou sXdaaova xov ZA, 
6 8s AZ xov AH ^.STpwv Xsitcstco sauTou eXdaaova xov HT, 
6 8s HT xov Z9 ^texpwv Xsitistw ^tovdSa xf]v 9A. 

Tksl ouv 6 E xov TA jiexpel, 6 8s TA xov BZ uexpel, 
xdi 6 E dpa xov BZ ^.expel' jiexpel 8s xdi oXov xov BA' 
xal Xoitcov dpa xov AZ [leiprpei.. 6 8s AZ xov AH ^isxpsT' 
xal 6 E dpa xov AH jisxpsT' [isxpsT 8s xal oXov xov AT- 
xal Xoittov dpa xov TH ^sxprjasi. 6 8s TH xov Z6 jisxpsl- 



B 1 1 D 

For two [unequal] numbers, AB and CD, the lesser 
being continually subtracted, in turn, from the greater, 
let the remainder never measure the (number) preceding 
it, until a unit remains. I say that AB and CD are prime 
to one another — that is to say, that a unit alone measures 
(both) AB and CD. 

For if AB and CD are not prime to one another then 
some number will measure them. Let (some number) 
measure them, and let it be E. And let CD measuring 
BF leave FA less than itself, and let AF measuring DC 
leave GC less than itself, and let GC measuring FH leave 
a unit, HA. 

In fact, since E measures CD, and CD measures BF, 
E thus also measures BF.^ And (E) also measures the 
whole of BA. Thus, (E) will also measure the remainder 



195 



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xal 6 E dpa xov Z6 jiexpeT- (jiexpel Be xal oXov xov ZA- ^li* 1 .* And AF measures DG. Thus, _E also measures DG. 
xal Xoikt]v apa xrjv A0 piovdBa ^exprjaei dpi%6<; wv oTtep And (i?) also measures the whole of DC. Thus, (E) will 
eaxlv dSuvaxov. oux dpa xou<; AB, TA dpi%ou<; {LSTpfpei also measure the remainder CG. And CG measures FH. 
xic dpi%6c oi AB, TA dpa npoxoi Kpoc dXXr]Xou<; eiaiv Thus, _E also measures FiJ. And also measures the 
oiiep eSei SeT^ai. whole of FA. Thus, (i?) will also measure the remaining 

unit AH, (despite) being a number. The very thing is 
impossible. Thus, some number does not measure (both) 
the numbers AB and CD. Thus, AB and CD are prime 
to one another. (Which is) the very thing it was required 
to show. 

t Here, use is made of the unstated common notion that if a measures b, and b measures c, then a also measures c, where all symbols denote 
numbers. 

t Here, use is made of the unstated common notion that if a measures b, and a measures part of b, then a also measures the remainder of b, where 
all symbols denote numbers. 



P'- 

Auo dpnJ^icov Bo'devxov \xr\ upcoxcov upoc; dXXr)Xouc; xo 
tieyiaxov auxov xoivov jiexpov eupelv. 



A 
E T 



B 1 



T r 
z 



H 



1 A 1 

"Eaxwaav oi Scedevxe^ Suo dprd^oi ^ir) npcoxoi Ttpoc 
dXXiqXouc; oi AB, TA. 8s! Br) x«v AB, TA xo ^eyiaxov xoivov 
(jiexpov eupe1v. 

Ei y.sv ouv 6 TA xov AB ^expet, ^expel Be xal eauxov, 6 
TA dpa xwv TA, AB xoivov ^Jtexpov eaxiv. xal cpavepov, oxi 
xai ^iiyiaxov ouBelc; yap ^eiCcov xou TA xov TA \LSTpr\oe\.. 

Ei Be ou [icxpel 6 TA xov AB, xwv AB, TA dvducpai- 
poujievou del xou eXdaaovo<; olko xou ^.eiCovoc; Xeicp , dr]aexai 
uc, dpid^toc;, bz \xexpr\oei xov upo eauxou. ^.ovac fiev 
yap ou Xeicp-driaexar ei Be (jltq, eaovxai oi AB, TA Kpwxoi 
7ip6<; dXXrjXouc ouep oux urcoxeixai. XeicpiSriaexai tic; dpa 
dpid^icx;, be, ^exprjoei xov Tipo eauxou. xai 6 \iev EA xov 
BE jiexpwv Xeiitexw eauxou eXdaaova xov EA, 6 Be EA xov 
AZ ^.expwv Xentexco eauxou eXdaaova xov ZT, 6 Be TZ xov 
AE [lexpeixw. euel ouv 6 TZ xov AE ^icxpel, 6 Be AE xov 
AZ ^lexpeT, xal 6 TZ apa xov AZ ^lexprjaei. [icxpel Be xal 
eauxov xal oXov dpa xov TA (jiexprpei. 6 Be TA xov BE 
^exper xal 6 TZ dpa xov BE ^texpeT- [icxpel Be xal xov EA- 
xal oXov apa xov BA [iexpr|aer ^texpeT Be xal xov TA- 6 TZ 
apa xou<; AB, EA ^texpeT. 6 TZ dpa xwv AB, TA xoivov 



Proposition 2 

To find the greatest common measure of two given 
numbers (which are) not prime to one another. 

A 



E - 



tC 
F 



Let AB and CD be the two given numbers (which 
are) not prime to one another. So it is required to find 
the greatest common measure of AB and CD. 

In fact, if CD measures AB, CD is thus a common 
measure oi CD and AB, (since CD) also measures itself. 
And (it is) manifest that (it is) also the greatest (com- 
mon measure). For nothing greater than CD can mea- 
sure CD. 

But if CD does not measure AB then some number 
will remain from AB and CD, the lesser being contin- 
ually subtracted, in turn, from the greater, which will 
measure the (number) preceding it. For a unit will not be 
left. But if not, AB and CD will be prime to one another 
[Prop. 7.1]. The very opposite thing was assumed. Thus, 
some number will remain which will measure the (num- 
ber) preceding it. And let CD measuring BE leave EA 
less than itself, and let EA measuring DF leave FC less 
than itself, and let CF measure AE. Therefore, since CF 
measures AE, and AE measures DF, CF will thus also 
measure DF. And it also measures itself. Thus, it will 



196 



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ELEMENTS BOOK 7 



[iexpov eaxiv. Xeyw 8r|, oxi xal [ieyiaxov. si yap [jlttj eaxiv 6 
rZ xfiv AB, TA [ieyiaxov xoivov [iexpov, [iexprpei tic; xouc; 
AB, TA dpii9^ouc; dpid[i6c; [iei^iov «v xou TZ. [iexpeixw, 
xal cgxgj 6 H. xal eitel 6 H xov TA [iexpeT, 6 6e TA xov 
BE [iexpel, xal 6 H dpa xov BE [iexpeT- [iexpeT 8e xal oXov 
xov BA- xal Xoinov apa xov AE [iexprjaei. 6 Se AE xov 
AZ [iexpeT- xal 6 H apa xov AZ [iexprpei- [iexpeT 8s xal 
oXov xov AE xal Xoittov apa xov TZ [iexpiqaei 6 [ieii^cov 
xov eXdaaova- 6%ep eaxlv dSuvaxov oux dpa xouc; AB, TA 
dpn5[iouc; dpidjioc; xu; [iexprpei [iei^wv wv xou TZ- 6 TZ apa 
xwv AB, TA [ieyiaxov £° Tt xoivov (iexpov [onep eSel Belial]. 



also measure the whole of CD. And CD measures BE. 
Thus, CF also measures BE. And it also measures EA. 
Thus, it will also measure the whole of BA. And it also 
measures CD. Thus, CF measures (both) AB and CD. 
Thus, CF is a common measure of AB and CD. So I say 
that (it is) also the greatest (common measure). For if 
CF is not the greatest common measure of AB and CD 
then some number which is greater than CF will mea- 
sure the numbers AB and CD. Let it (so) measure (AB 
and CD), and let it be G. And since G measures CD, 
and CD measures BE, G thus also measures BE. And it 
also measures the whole of BA. Thus, it will also mea- 
sure the remainder AE. And AE measures DF. Thus, G 
will also measure DF. And it also measures the whole 
of DC. Thus, it will also measure the remainder CF, 
the greater (measuring) the lesser. The very thing is im- 
possible. Thus, some number which is greater than CF 
cannot measure the numbers AB and CD. Thus, CF is 
the greatest common measure of AB and CD. [(Which 
is) the very thing it was required to show] . 



I16pia[jia. 

'Ex 8r) xouxou cpavepov, oxi edv dpid[i6c; 8uo dpid[iouc; 
[iexprj, xal xo [ieyiaxov auxSSv xoivov [iexpov [iexprpei- ojtep 
e8ei 8eT^ai. 



Corollary 

So it is manifest, from this, that if a number measures 
two numbers then it will also measure their greatest com- 
mon measure. (Which is) the very thing it was required 
to show. 



T • 

Tpiwv dpid[ic5v So'devxwv [irj Ttpwxcov izpoc, dXXr|Xouc; to 
[ieyiaxov auxov xoivov [iexpov eupeTv. 



Proposition 3 

To find the greatest common measure of three given 
numbers (which are) not prime to one another. 



A B T A E Z 

TCaxoaav oi Bo-devxec; xpeT? dpn9[iol ]if\ Tipoxoi TCpoc 
dXXf|Xouc; oi A, B, E 5eT 8rj xov A, B, T xo [ieyiaxov xoivov 
[iexpov eupeTv. 

EiXr](p'd« yap Suo xwv A, B xo [ieyiaxov xoivov [iexpov 6 
A- 6 8r) A xov T rjxoi [iexpeT rj ou [iexpeT. [iexpeixw itpoxepov 
[iexpeT Se xal xouc; A, B- 6 A apa xouc; A, B, T [iexpeT- 6 
A apa xwv A, B, T xoivov [iexpov eaxiv. Xeyw 5r|, oxi xal 



A B C D E F 

Let A, B, and C be the three given numbers (which 
are) not prime to one another. So it is required to find 
the greatest common measure of A, B, and C. 

For let the greatest common measure, D, of the two 
(numbers) A and B have been taken [Prop. 7.2]. So D 
either measures, or does not measure, C. First of all, let 
it measure (C). And it also measures A and B. Thus, D 



197 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



(ifyioTov. si yap ^Vj eoxiv 6 A xwv A, B, T ^(iyiaTov xoivov 
^tETpov, ^exprjaei tic; touc; A, B, T dpid^touc; dpidfjioc; ^eic^cov 
wv tou A. jiexpeixw, xal eotco 6 E. etc! ouv 6 E touc; A, B, 
r [lexpsT, xal touc; A, B apa \iETprpe^ xal to twv A, B dpa 
^liyioTov xoivov (jiETpov \xeTpf\aei. to 8s tGv A, B jieyiaTov 
xoivov ^iSTpov scttIv 6 A- 6 E apa tov A [iETpsT 6 ^sii^cov 
tov eXdaaova - ouep eotiv dSuvaTov. oux apa touc; A, B, T 
dpid^touc; dpi-djioc; tic; \xexpr\oei ^.aCtov wv tou A- 6 A apa 
twv A, B, T jieyioTov ecru xoivov ^STpov. 

Mf) [iSTpeiTW 8r) 6 A tov E XsyM TtpfiTov, oti oi T, A 
oux siai KpwToi Tipoc; dXXr|Xou<;. excel yap oi A, B, T oux 
eiai KptOTOi Tipoc dXXr|Xouc;, [iSTpr|aei tic; auTouc; dpi%6c;. 6 
8r) touc; A, B, T ^.ETpwv xal touc; A, B |i£Tpr|aei, xal to 
twv A, B ^iyicrcov xoivov jieTpov tov A \iSTpr\ae\.- ^.ETpei 
8s xal tov T- touc; A, T apa dpi%ouc; dpi-d^oc; tic; ^STpr]oei- 
oi A, r apa oux eiai TipoToi Ttpoc; dXXr|Xouc;. eiXricpiDw ouv 
auTOv to ^syiaTov xoivov jiSTpov 6 E. xal end 6 E tov A 
[LejpeT, 6 Se A touc; A, B ^ETpeT, xal 6 E apa touc; A, B 
^.ETpeT - ^(.ETpei 8e xal tov E 6 E apa touc; A, B, T (iSTpeT. 
6 E apa iSv A, B, T xoivov ecru (iETpov. Xeyco br\, oti xal 
^eyiaTov. si yap u/] ecruv 6 E t£Sv A, B, T to ^leyiaTov 
xoivov [iSTpov, \ieTpr\aei tic; touc; A, B, T dpid^ouc; dpnD^toc; 
^.EiCwv <2>v tou E. [iSTpeiTO, xal sgto 6 Z. xal ETtd 6 Z touc; 
A, B, r ^iSTpei, xal touc; A, B ^.ETpei' xal to twv A, B apa 
^iiyiaTov xoivov [iSTpov \iexp'f]aei. to Se t£>v A, B ^eyiaTOv 
xoivov ^.ETpov sotIv 6 A- 6 Z dpa tov A \isxpev [izxpei 8s 
xal tov T- 6 Z apa touc; A, T ^i£Tper xal to twv A, T apa 
^iiyiaTov xoivov (iSTpov \LeTpf\aei. to 8e tcjv A, T jieyiaTOv 
xoivov [iSTpov eotIv 6 E - 6 Z apa tov E [lexpei 6 [le'i^v 
tov eXdaaova- ouep eotIv dSuvaTov. oux apa touc; A, B, T 
dpid^touc; dpid^ioc; tic; \iexpr\oei jieiCwv wv tou E - 6 E apa 
tcjv A, B, r jieyiaTov eoti xoivov ^ETpov onep e8ei Seic;ai. 



measures A, B, and C. Thus, D is a common measure 
of A, B, and C. So I say that (it is) also the greatest 
(common measure) . For if D is not the greatest common 
measure of A, B, and C then some number greater than 
D will measure the numbers A, B, and C. Let it (so) 
measure {A, B, and C), and let it be E. Therefore, since 
E measures A, B, and C, it will thus also measure A and 
B. Thus, it will also measure the greatest common mea- 
sure of A and B [Prop. 7.2 corr.]. And D is the greatest 
common measure of A and B. Thus, E measures D, the 
greater (measuring) the lesser. The very thing is impossi- 
ble. Thus, some number which is greater than D cannot 
measure the numbers A, B, and C. Thus, D is the great- 
est common measure of A, B, and C. 

So let D not measure C. I say, first of all, that C 
and D are not prime to one another. For since A, B, C 
are not prime to one another, some number will measure 
them. So the (number) measuring A, B, and C will also 
measure A and B, and it will also measure the greatest 
common measure, D, of A and B [Prop. 7.2 corr.]. And 
it also measures C. Thus, some number will measure the 
numbers D and C. Thus, D and C are not prime to one 
another. Therefore, let their greatest common measure, 
E, have been taken [Prop. 7.2] . And since E measures 

D, and D measures A and B, E thus also measures A 
and B. And it also measures C. Thus, E measures A, B, 
and C. Thus, E is a common measure of A, B, and C. So 
I say that (it is) also the greatest (common measure). For 
if E is not the greatest common measure of A, B, and C 
then some number greater than E will measure the num- 
bers A, B, and C. Let it (so) measure (A, B, and C), and 
let it be F. And since F measures A, B, and C, it also 
measures A and B. Thus, it will also measure the great- 
est common measure of A and B [Prop. 7.2 corr.]. And 
D is the greatest common measure of A and B. Thus, F 
measures D. And it also measures C. Thus, F measures 
D and C. Thus, it will also measure the greatest com- 
mon measure of D and C [Prop. 7.2 corr.]. And E is the 
greatest common measure of D and C. Thus, F measures 

E, the greater (measuring) the lesser. The very thing is 
impossible. Thus, some number which is greater than E 
does not measure the numbers A, B, and C. Thus, E is 
the greatest common measure of A, B, and C. (Which 
is) the very thing it was required to show. 



"Auac; dpidjioc; navToc; dpiduou 6 eXdaawv tou ^id^ovoc; 
f]TOi [Jispoc; egtIv fj [L&pr\. 

'EaTwaav 8uo dpi$uol oi A, BT, xal scttm eXdaawv 6 
Br- Xeyw, oti 6 BT tou A j^toi ^lepoc; e<xclv fj [lept]. 



Proposition 4 

Any number is either part or parts of any (other) num- 
ber, the lesser of the greater. 

Let A and BC be two numbers, and let BC be the 
lesser. I say that BC is either part or parts of A. 



198 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



Oi A, Br yap fjxoi Ttpaixoi npbc, dXXrjXouc; eiaiv f] ou. 
eoxcooav rcpoxepov oi A, Br Ttpfixoi upog dXXiqXouc;. Biai- 
pei&£VToc; 8/) xou Br el? iac, ev auxfi ^tovdBa<; eoxai exdoxr) 
[iovac x«v sv xG Br [izpoc, xi xoO A- wax£ (iepr) eoxiv 6 Br 
xou A. 



For A and BC are either prime to one another, or not. 
Let A and BC, first of all, be prime to one another. So 
separating BC into its constituent units, each of the units 
in BC will be some part of A. Hence, BC is parts of A. 



Bi 



B T 



A 



Mr] eoxcooav Sr| oi A, Br upcoxoi 7ip6<; dXXrjXout;' 6 8r) 
Br xov A fjxoi [iztpzi fj ou (jiexpel. si [ie\ ouv 6 Br xov 
A ^lexpeT, [ispoz eoxiv 6 Br xoO A. ei 8e ou, eiAfjcpi&M xfiv 
A, Br [icyioxov xoivov [iexpov 6 A, xai 8ir]piqoiL>M 6 Br si<z 
xou? xG A Taoue; xoug BE, EZ, Zr. xai enei 6 A xov A 
(jiexpsi, [izpoc, eoxiv 6 A xou A- i'ooc; 8e 6 A exdoxip xfiv 
BE, EZ, Zr- xai exaoxoc; dpa xfiv BE, EZ, Zr xou A ^epoc 
eoxiv a>oxe [ispi] eoxiv 6 Br xou A. 

"Anotz dpa dpiduoc; Tiavxog api^ou 6 eXdoowv xou 
(jieiCovog fjxoi ^epoc; eoxiv fj piepr)- onep eBei 8el^ai. 



A D 

So let A and BC be not prime to one another. So BC 
either measures, or does not measure, A. Therefore, if 
BC measures A then BC is part of A. And if not, let the 
greatest common measure, D, of A and BC have been 
taken [Prop. 7.2], and let BC have been divided into BE, 
EF, and FC, equal to D. And since D measures A, D is 
a part of A And D is equal to each of BE, EF, and FC. 
Thus, BE, EF, and FC are also each part of A. Hence, 
BC is parts of A. 

Thus, any number is either part or parts of any (other) 
number, the lesser of the greater. (Which is) the very 
thing it was required to show. 



Proposition 5* 



'Edv dpii^oc; dpi^ou ^epo<; fj, xai exepoc; exepou xo 
auxo \xepoc, fj, xai ouvajicpoxepoc; ouva^cpoxepou xo auxo 
[iepoc; eoxai, onep 6 elg xou evoc;. 



B 



H 



r 



T E 



z 



A A 

Apiduoc; yap 6 A [dpiduou] xou Br ^tepoc; eaxw, xai 



If a number is part of a number, and another (num- 
ber) is the same part of another, then the sum (of the 
leading numbers) will also be the same part of the sum 
(of the following numbers) that one (number) is of an- 
other. 



B 



G 



tE 



H 



A D 

For let a number A be part of a [number] BC, and 



199 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



exepoc; 6 A exepou xou EZ to auxo ^tepoc;, ojtep 6 A xou 
Br- Xeyo, oxi xai auva^tcpoxepoc; 6 A, A auvajicpoxepou xou 
Br, EZ xo auxo \iepoc, eaxiv, ojtep 6 A xou Br. 

'Etc! yap, o jiepoc; eaxlv 6 A xou Br, xo auxo ^tepoc; eaxl 
xai 6 A xou EZ, oaoi apa eialv ev xfii Br dpi-d^ol tool iu 
A, xoaouxoi eiai xai ev xQ EZ dpid^tol taoi xw A. 5ifjpr]C7d« 
6 [Lev Br Etc; xouc; x£5 A taouc; xouc; BH, Hr, 6 8e EZ eic; 
xouc; xo A i'aouc; xouc; EO, 8Z- eaxai 8r) taov xo iiXfj'doc; 
xov BH, Hr xw TtXyydei xwv E6, 8Z. xai £7tei taoc; eaxlv 
6 \ik\ BH xw A, 6 8e E8 xw A, xai oi BH, EG dpa xoTc; 
A, A tool. Sid xd auxd 8f] xai oi Hr, 0Z xoTc A, A. oaoi 
dpa [eialv] ev xo Br dpid^tol tool xo A, xoaouxoi rial xai 
ev xoTc; Br, EZ tool xou; A, A. oaaitXaaiwv apa eaxlv 6 Br 
xou A, xoaauxaTiXaaiov eaxl xai auvajicpoxepoc; 6 Br, EZ 
auva^tcpoxepou xou A, A. o apa ^tepoc; eaxlv 6 A xou Br, xo 
auxo [lepoc, eaxl xai auvajicpoxepoc; 6 A, A auva^icpoxepou 
xou Br, EZ- oTtep e8ei SeTc;ai. 



another (number) D (be) the same part of another (num- 
ber) EF that A (is) of BC. I say that the sum A, D is also 
the same part of the sum BC, EF that A (is) of BC. 

For since which(ever) part A is of BC, D is the same 
part of EF, thus as many numbers as are in BC equal 
to A, so many numbers are also in EF equal to D. Let 
BC have been divided into BG and GC, equal to A, and 
EF into EH and HF, equal to D. So the multitude of 
(divisions) BG, GC will be equal to the multitude of (di- 
visions) EH, HF. And since BG is equal to A, and EH 
to D, thus BG, EH (is) also equal to A, D. So, for the 
same (reasons), GC, HF (is) also (equal) to A, D. Thus, 
as many numbers as [are] in BC equal to A, so many are 
also in BC, EF equal to A, D. Thus, as many times as 
BC is (divisible) by A, so many times is the sum BC, EF 
also (divisible) by the sum A, D. Thus, which(ever) part 
A is of BC, the sum A, D is also the same part of the 
sum BC, EF. (Which is) the very thing it was required 
to show. 



t In modem notation, this proposition states that if a = (l/n)b and c = (1/n) d then (a + c) = (1/n) (6 + d), where all symbols denote numbers. 



Proposition 6 f 



'Edv dpid^ioc; dpi'djiou (iepr) fj, xai exepoc; exepou xa 
auxd [iepr] fj, xai auvajicpoxepoc; auvapicpoxepou xa auxd 
[ispr] eaxai, onep 6 eic; xou evoc;. 



A 
H 



1 B 1 

r 



A 



iE z 



Apidjioc; yap 6 AB dpiiD^tou xou T [icpr] eaxw, xai exepoc; 
6 AE exepou xou Z xa auxd ^epr), anep 6 AB xou T- Xeyto, 
oxi xai auva^xcpoxepoc; 6 AB, AE auva^xcpoxepou xou T, Z 
xd auxd \iipr\ eaxiv, drcep 6 AB xou T. 

Tkel yap, a [iepf] eaxlv 6 AB xou T, xd auxd [iepr\ xai 
6 AE xou Z, oaa dpa eaxlv ev xw AB [iepf] xou T, xoaauxd 
eaxi xai ev xw AE [ispf] xou Z. 8ir]pr]a'f)w 6 [Lev AB eic; xd 
xou T \iept) xd AH, HB, 6 8e AE eic; xd xou Z [iepf] xd 
A0, 6E- eaxai 8r] laov xo TtXrydoc; x«v AH, HB xw txXt)tl>ei 
xwv A0, 6E. xai etxel, o uepoc; eaxlv 6 AH xou T, xo 



If a number is parts of a number, and another (num- 
ber) is the same parts of another, then the sum (of the 
leading numbers) will also be the same parts of the sum 
(of the following numbers) that one (number) is of an- 
other. 



A 
G 
B 



D 
H 



For let a number AB be parts of a number C, and an- 
other (number) DE (be) the same parts of another (num- 
ber) F that AB (is) of C. I say that the sum AB, DE is 
also the same parts of the sum C, F that AB (is) of C. 

For since which (ever) parts AB is of C, DE (is) also 
the same parts of F, thus as many parts of C as are in AB, 
so many parts of F are also in DE. Let AB have been 
divided into the parts of C, AG and GB, and DE into the 
parts of F, DH and HE. So the multitude of (divisions) 
AG, GB will be equal to the multitude of (divisions) DH, 



200 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



auxo ^epoc; eaxl xai 6 A0 xou Z, 8 dpa \±spoz eaxlv 6 AH 
xoO T, to auxo \Lspoz eaxl xai auva|icp6xepo<; 6 AH, AO 
auva^tcpoxepou xou T, Z. Sid xa auxa 8f] xai 6 ^epo<; eaxlv 6 
HB xoO r, xo auxo ^epoc eaxl xai auva^cpoxepo<; 6 HB, 9E 
auva^icpoxepou xou T, Z. a apa ^epr) eaxlv 6 AB xoO T, xa 
auxa (jiepr) eaxl xai auva^icpoxepog 6 AB, AE auva^tcpoxepou 
xou T, Z- ouep eBei 8el£ai. 



And since which (ever) part AG is of C, DH is also 
the same part of F, thus which(ever) part AG is of C, 
the sum AG, DH is also the same part of the sum C, F 
[Prop. 7.5]. And so, for the same (reasons), which(ever) 
part GB is of C, the sum GB, HE is also the same part 
of the sum C, F. Thus, which (ever) parts AB is of C, 
the sum AB, DE is also the same parts of the sum C, F. 
(Which is) the very thing it was required to show. 



t In modem notation, this proposition states that if a = (m/n) b and c = (m/n) d then (a + c) = (m/n) (b + d), where all symbols denote 
numbers. 



'Edv dpL-djioc; dpi-djiou [icpoc fj, onep dcpaipe-dslc; dcpai- 
pei^evxoc;, xai 6 Xoltt6<; xou Xoittou xo auxo [lepoc. eaxai, 
OTtep 6 oXoc; xou oXou. 

A E B 

i — i 1 

H r Z A 

I 1 1 1 

Apii9(ji6(; jap 6 AB dpi-fijiou xou IA |iepo<; eaxo, oitep 
dcpaips-dric; 6 AE dcpaipeiJevxoc; xou TZ- Xeyw, oxi xai Xomog 
6 EB Xoittou xou ZA xo auxo y.epoc, eoxiv, oTisp oXoc; 6 AB 
oXou xou TA. 

"0 yap [lepoz iaxlv 6 AE xou TZ, xo auxo [izpoQ eaxco 
xai 6 EB xou TH. xai ine'i, o ^.epoc; saxlv 6 AE xou TZ, xo 
auxo \xspoz eaxl xai 6 EB xou TH, o apa ^.epo<; eaxlv 6 AE 
xou TZ, xo auxo ^iepo<; eaxl xai 6 AB xou HZ. o Be ^tepo<; 
eaxlv 6 AE xou TZ, xo auxo \±spoz (moxeixai xai 6 AB xou 
TA- o apa \±epoq eaxl xai 6 AB xou HZ, xo auxo \±epoq eaxl 
xai xou TA- l'ao<; apa eaxlv 6 HZ ifi TA. xoivo<; dtpr)pr]a , dw 
6 TZ' Xoikoz apa 6 Hr Xomw xw ZA eaxiv Xaoc,. xai etxeC, 
o ^.epoc eaxlv 6 AE xou TZ, xo auxo (icpog [eaxl] xai 6 EB 
xou Hr, Xaoc, 8e 6 Hr iu ZA, S apa ^iepo<; eaxlv 6 AE xou 
TZ, xo auxo |iepo<; eaxl xai 6 EB xou ZA. dXXa o ^tepo<; 
eaxlv 6 AE xou TZ, xo auxo ^.epoc; eaxl xai 6 AB xou FA- 
xai Xom6<; apa 6 EB Xomou xou ZA xo auxo ^tepo<; eaxlv, 
oTiep oXoc, 6 AB oXou xou TA- onep e8ei SeT^ai. 



Proposition 7 f 

If a number is that part of a number that a (part) 
taken away (is) of a (part) taken away then the remain- 
der will also be the same part of the remainder that the 
whole (is) of the whole. 

A E B 

i — i 1 

G C F D 

i 1 1 1 

For let a number AB be that part of a number CD 
that a (part) taken away AE (is) of a part taken away 
CF. I say that the remainder EB is also the same part of 
the remainder FD that the whole AB (is) of the whole 
CD. 

For which(ever) part AE is of CF, let EB also be the 
same part of CG. And since which(ever) part AE is of 
CF, EB is also the same part of CG, thus which(ever) 
part AE is of CF, AB is also the same part of GF 
[Prop. 7.5]. And which(ever) part AE is of CF, AB is 
also assumed (to be) the same part of CD. Thus, also, 
which (ever) part AB is of GF, (AB) is also the same 
part of CD. Thus, GF is equal to CD. Let CF have been 
subtracted from both. Thus, the remainder GC is equal 
to the remainder FD. And since which (ever) part AE is 
of CF, EB [is] also the same part of GC, and GC (is) 
equal to FD, thus which(ever) part AE is of CF, EB is 
also the same part of FD. But, which(ever) part AE is of 
CF, AB is also the same part of CD. Thus, the remain- 
der EB is also the same part of the remainder FD that 
the whole AB (is) of the whole CD. (Which is) the very 
thing it was required to show. 



t In modern notation, this proposition states that if a = (1/n) b and c = (1/n) d then (a — c) = (1/n) (6— d), where all symbols denote numbers. 



7]'. Proposition 8 f 

'Eav dpidjioc; dpidjiou \ispt) rj, drcep dcpaipeiitelc; dcpai- If a number is those parts of a number that a (part) 

pe$evxo<;, xai 6 Xolttoc xou Xoittou xa auxa [iepr] eaxai, taken away (is) of a (part) taken away then the remain- 

ditep 6 8Xo<; xou oXou. der will also be the same parts of the remainder that the 



201 



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ELEMENTS BOOK 7 



whole (is) of the whole. 

r Z A C F D 

I 1 1 I 1 1 

H M K N GMKNH 

i 1 — i 1 — i i 1 — i 1 — i 



A A E B 

i 1 1 1 

Apid^io? yap o AB dpi/d^iou tou IA \±epr\ screw, abtep 
acpaipsdsle; 6 AE dcpaipsiSsvToc; tou TZ- Xsyto, oxi xal Xomoc 
6 EB Xoittou xou ZA Ta auTa [ispf] sot(v, arcsp oXoc; 6 AB 
oXou tou TA. 

Ksitxdo yap tG AB iaoc 6 H8, a dpa \±£pr] scrav 6 H9 
tou TA, Ta ai)Ta ^ispr] sera xal 6 AE tou TZ. 8ir)pr]a , do 6 
^isv H9 si? Ta tou TA [leprj Ta HK, K9, 6 Ss AE tic, Ta tou 
TZ [itpr\ Ta AA, AE- saTai Sr] laov to TtXrydoc; t«v HK, K0 
Tip TcXrjn&ei tcov AA, AE. xal inti, 6 ^epog scrav 6 HK tou 
TA, to auTo y.epoc sera xal 6 AA tou TZ, jis^tov 8s 6 TA 
tou TZ, [isi^tov apa xal 6 HK tou AA. xsicrdio Ttp AA faoc; 
6 HM. o apa ^ispoc; saxlv 6 HK tou TA, to auTo ^ispoc ecra 
xal 6 HM tou TZ- xal Xoinbz apa 6 MK XomoO tou ZA 
to auTo ^epoc sotiv, OTtsp oXoc; 6 HK oXou tou TA. TtdXiv 
stcsl, o fispoc scrav 6 K9 tou TA, to auTo ^ispo<; sera xal 6 
EA tou TZ, [idCwv Ss 6 TA tou TZ, ^si^cov dpa xal 6 6K 
tou EA. xsicrdco to EA taoc. 6 KN. o apa ^tspo<; scrav 6 K0 
tou TA, to auTo ^tepo<; sera xal 6 KN tou TZ- xal Xoitcoc 
dpa 6 N9 Xoitiou tou ZA to auTo [izpoc, scrav, oitsp oXo<; 6 
K9 oXou tou TA. sBsix'dr) 8s xal Xomo<; 6 MK Xoitiou tou 
ZA to ai)To ^tepot; <J>v, OTisp okoc 6 HK oXou tou TA- xal 
auva^tcpoTspoc; apa 6 MK, N9 tou AZ Ta auTa [itpf] scrav, 
djtsp okoc, 6 9H oXou tou TA. Taoc, 8s auva^tcpoTspoc [isv 
6 MK, N9 tw EB, 6 Bs 9H to BA- xal Xomoc. dpa 6 EB 
Xoitiou tou ZA Ta auTa [itpf] scrav, djisp 6Xoc. 6 AB oXou 
tou IA- oitsp eBsi 8su;ai. 



A L E B 

i 1 1 1 

For let a number AB be those parts of a number CD 
that a (part) taken away AE (is) of a (part) taken away 
CF. I say that the remainder EB is also the same parts 
of the remainder FD that the whole AB (is) of the whole 
CD. 

For let GH be laid down equal to AB. Thus, 
which (ever) parts GH is of CD, AE is also the same 
parts of CF. Let GH have been divided into the parts 
of CD, GK and KH, and AE into the part of CF, AL 
and LE. So the multitude of (divisions) GK, KH will be 
equal to the multitude of (divisions) AL, LE. And since 
which (ever) part GK is of CD, AL is also the same part 
of CF, and CD (is) greater than CF, GK (is) thus also 
greater than AL. Let GM be made equal to AL. Thus, 
which(ever) part GK is of CD, GM is also the same part 
of CF. Thus, the remainder MK is also the same part of 
the remainder FD that the whole GK (is) of the whole 
CD [Prop. 7.5]. Again, since which(ever) part KH is of 
CD, EL is also the same part of CF, and CD (is) greater 
than CF, HK (is) thus also greater than EL. Let KN be 
made equal to EL. Thus, which(ever) part KH (is) of 
CD, KN is also the same part of CF. Thus, the remain- 
der NH is also the same part of the remainder FD that 
the whole KH (is) of the whole CD [Prop. 7.5]. And the 
remainder MK was also shown to be the same part of 
the remainder FD that the whole GK (is) of the whole 
CD. Thus, the sum MK, NH is the same parts of DF 
that the whole HG (is) of the whole CD. And the sum 
MK, NH (is) equal to EB, and HG to BA. Thus, the 
remainder EB is also the same parts of the remainder 
FD that the whole AB (is) of the whole CD. (Which is) 
the very thing it was required to show. 



t In modem notation, this proposition states that if a = (m/n)b and c = (m/n) d then (a - c) = (m/n) (b - d), where all symbols denote 
numbers. 

Proposition 9 1 

'Edv dpi%6c dpi%ou [lepoc ?j, xal STspog STspou to If a number is part of a number, and another (num- 
auTo ^spoc; fj, xal svaXXdc;, o ^spoc; scrav fj (ispr) 6 np&Toc ber) is the same part of another, also, alternately, 
tou Tphou, to auTo jjispoc. saTai f\ Ta auTa [iepr\ xal 6 which(ever) part, or parts, the first (number) is of the 
BsuTspoc. tou TETdpTou. third, the second (number) will also be the same part, or 



202 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



B 



H 



A 



A 



Apidjioc; yap 6 A dpidjioO xou Br \±spoz eaxo, xal exe- 
poc 6 A exepou xou EZ to auxo [iepo<;, ojtep 6 A xou Br^ 
Xeyw, oxi xai evaXXd?, o \iepoq eaxlv 6 A xou A fj ^epr), xo 
auxo \xepoc, eaxl xal 6 Br xou EZ f] ^epr). 

'End yap o ^spo<; eaxlv 6 A xou Br, xo auxo ^.epoc eaxl 
xal 6 A xou EZ, oaoi apa eialv ev iw Br dpid^ol laoi x£> 
A, TooouTo'i eiai xal ev xw EZ Taoi xc5 A. 8iy]pfja , d« 6 [lev 
Br tic, xou? xc5 A laouc xou<; BH, Hr, 6 Se EZ el? xou<; iu 
A laouc; xou<; E6, 0Z- eaxai 8f] laov xo TiXfjiEkK xGv BH, 
Hr iS TiXVjflei xwv EG, ez. 

Kai ekei Taoi elalv oi BH, Hr apid^iol dXXfjXoic;, eial 
8e xal oi EG, 0Z dpi/d^ol i'aoi dXXfjXou;, xa[ eaxiv i'aov xo 
TiXfj'dot; iwv BH, Hr xCS irXfj'dei xwv E9, 0Z, 8 dpa y.epoc, 
eaxlv 6 BH xou E9 fj [iepf], xo auxo ^tepoc; eaxl xal 6 Hr 
xou 9Z fj xd auxa \i£pr\ % &axe xal 6 ^iepo<; eaxlv 6 BH xou 
E9 fj [i£pr\, xo auxo ^tepo<; eaxl xal auva^icp6xepo<; 6 Br 
auva^tcpoxepou xou EZ fj xd auxa [ispr\. Xooc, Se 6 uev BH 
tu A, 6 8e E9 x£> A' o apa ^tepo<; eaxlv 6 A xou A fj ^epr], 
xo auxo ^lepoc eaxl xal 6 Br xou EZ fj xd auxa [i£pr\- onep 
e8ei 8el^ai. 



the same parts, of the fourth. 



B 



E T 



H 



A D 

For let a number A be part of a number BC, and an- 
other (number) D (be) the same part of another EF that 
A (is) of BC. I say that, also, alternately, which(ever) 
part, or parts, A is of D, BC is also the same part, or 
parts, of EF. 

For since which(ever) part A is of BC, D is also the 
same part of EF, thus as many numbers as are in BC 
equal to A, so many are also in EF equal to D. Let BC 
have been divided into BC and GC, equal to A, and EF 
into -EiJ and HF, equal to £>. So the multitude of (di- 
visions) BC, GC will be equal to the multitude of (divi- 
sions) EH, HF. 

And since the numbers BC and GC are equal to one 
another, and the numbers EH and HF are also equal to 
one another, and the multitude of (divisions) BG, GC 
is equal to the multitude of (divisions) EH, HC, thus 
which (ever) part, or parts, BG is of EH, GC is also 
the same part, or the same parts, of HF. And hence, 
which (ever) part, or parts, BG is of EH, the sum BC 
is also the same part, or the same parts, of the sum EF 
[Props. 7.5, 7.6]. And BG (is) equal to A, and EH to D. 
Thus, which(ever) part, or parts, A is of D, BC is also 
the same part, or the same parts, of EF. (Which is) the 
very thing it was required to show. 



t In modern notation, this proposition states that if a 
numbers. 



(1/n) b and c = (1/n) d then if a = (k/l) c then b = (k/l) d, where all symbols denote 



I . 

'Edv dpiv}^6<; dpnf)[iou [ispf] fj, xal exepoc; exepou xd auxa 
[Jtepr] fj, xal evaXXd^, a \iepr\ eaxlv 6 npfixot; xou xpixou fj 
fjiepoc, xd auxa fiepr] eaxai xal 6 Beuxepoc xou xexdpxou fj 
xo auxo (iepoc- 

Api-djioi; yap 6 AB dpid^iou xou T \±epr) eaxco, xal exepoc; 
6 AE exepou xou Z xa auxa [iepf]- Xeyw, oxi xal evaXXd^, 
a [iepi] eaxlv 6 AB xou AE fj ^tepo<;, xa auxa [ispf] eaxl xal 
6 r xou Z fj xo auxo ^tepo<;. 



Proposition 10 f 

If a number is parts of a number, and another (num- 
ber) is the same parts of another, also, alternately, 
which (ever) parts, or part, the first (number) is of the 
third, the second will also be the same parts, or the same 
part, of the fourth. 

For let a number AB be parts of a number C, and 
another (number) DE (be) the same parts of another F. 
I say that, also, alternately, which (ever) parts, or part, 



203 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



AB is of DE, C is also the same parts, or the same part, 
of F. 



A 
H 
B 



r 



A 

® 

E 



D 
H 



'Etc! y^Pj & t^p/] eaxlv 6 AB xou T, xd auxa [ispr\ eaxl 
xal 6 AE xou Z, oaa apa eaxlv ev xw AB [i£pr) xoO T, 
xoaauxa xal ev xw AE ^ep/] xou Z. BupfjcTdw 6 [lev AB eu; 
xa xou r [-tepr) xa AH, HB, 6 8e AE eu; xa xou Z (jiepr) xa 
A0, 9E- eaxai 8f] laov xo TtXrji!)o<; ifiv AH, HB xo TcXyydei 
iwv A0, 0E. xal ETtei, o ^iepo<; eaxlv 6 AH xou T, xo auxo 
^iepo<; eaxl xal 6 A6 xou Z, xal evaXXd?;, o jiepo<; eaxlv 6 
AH xou A8 fj ^epr], xo auxo ^iepo<; eaxl xal 6 T xou Z fj 
xa auxa (ieprj. 8ia xa auxa Sf] xa(, 8 ^lipoc; eaxlv 6 HB xou 
6E fj [iepf] 7 xo auxo jiepo<; eaxl xal 6 T xou Z fj xa auxa 
^.epr]- waxe xal [8 jiepo<; eaxlv 6 AH xou A9 fj \±spr}, xo 
auxo ^iepo<; eaxl xal 6 HB xou 0E fj xa auxa ^epr)- xal o 
apa ^tepo<; eaxlv 6 AH xou AO fj ^epr], xo auxo uepo<; eaxl 
xal 6 AB xou AE fj xa auxa y.ept}- dXX' 8 ^tepo<; eaxlv 6 AH 
xou A0 fj [iepf], xo auxo [izpoc, eBelx^/] xal 6 T xou Z fj xa 
auxa [ispt], xal] a [dpa] ^iepr] eaxlv 6 AB xou AE fj [iepo<;, 
xa auxa jiepr) eaxl xal 6 T xou Z fj xo auxo jiepoc 8icep eSei 
Bel^ai. 



t In modem notation, this proposition states that if a : 
numbers. 



A 
G 
B 



For since which (ever) parts AB is of C, DE is also 
the same parts of F, thus as many parts of C as are in 
AB, so many parts of F (are) also in DE. Let AB have 
been divided into the parts of C, AG and GB, and DE 
into the parts of F, DH and ifi£. So the multitude of 
(divisions) AG, GB will be equal to the multitude of (di- 
visions) DH, HE. And since which (ever) part AG is 
of C, DH is also the same part of F, also, alternately, 
which (ever) part, or parts, AG is of DH, C is also the 
same part, or the same parts, of F [Prop. 7.9]. And so, 
for the same (reasons), which (ever) part, or parts, GB is 
of HE, C is also the same part, or the same parts, of F 
[Prop. 7.9]. And so [which(ever) part, or parts, AG is of 
DH, GB is also the same part, or the same parts, of HE. 
And thus, which(ever) part, or parts, AG is of DH, AB is 
also the same part, or the same parts, of DE [Props. 7.5, 
7.6]. But, which(ever) part, or parts, AG is of DH, C 
was also shown (to be) the same part, or the same parts, 
of F. And, thus] which (ever) parts, or part, AB is of DE, 
C is also the same parts, or the same part, of F. (Which 
is) the very thing it was required to show. 



(m/n)b and c = (m/n) d then if a = (k/l) c then b = (k/l) d, where all symbols denote 



ia . 

Eav fj toe; oXoc; npbz oXov, ouxcoc; dcpaipeiDelt; itpo<; dcpai- 
peiJevxa, xal 6 Xoittoc; upog xov Xomov eaxai, tbg oXoc; upog 
oXov. 

'Eaxa> obc; oXoc; 6 AB npoc; oXov xov TA, ouxoc dcpai- 
pe$el<; 6 AE npbc, dcpaipeOevxa xov TZ- Xeyto, oxi xal Xoitcoc 
6 EB Ttpoc Xomov xov ZA eaxiv, 6<; oXo<; 6 AB icpog oXov 

xov rA. 



Proposition 11 

If as the whole (of a number) is to the whole (of an- 
other), so a (part) taken away (is) to a (part) taken away, 
then the remainder will also be to the remainder as the 
whole (is) to the whole. 

Let the whole AB be to the whole CD as the (part) 
taken away AE (is) to the (part) taken away CF. I say 
that the remainder EB is to the remainder FD as the 
whole AB (is) to the whole CD. 



204 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



r 



A 
E 



B 1 A 

'Enei ecmv &c 6 AB npoc tov TA, outcoc 6 AE npoc 
tov rZ, 8 apa \±spo<z ecmv 6 AB tou TA f\ \xepr\, to auxo 
[ispoz soil xal 6 AE tou TZ fj xa aura [isprj. xal Xomoc 
apa 6 EB XomoO tou ZA to auTO ^iipoc eotIv f) (iepr), dnep 
6 AB tou TA. eaTiv apa (be 6 EB npoc tov ZA, outmc 6 
AB npoc; tov TA- onsp s8ei Bsicai. 



A 
E 



B 1 D J 

(For) since as AB is to CD, so AE (is) to CF, thus 
which(ever) part, or parts, AB is of CD, AE is also the 
same part, or the same parts, of CF [Def. 7.20]. Thus, 
the remainder EB is also the same part, or parts, of the 
remainder FD that AB (is) of CD [Props. 7.7, 7.8]. 
Thus, as EB is to FD, so AB (is) to CD [Def. 7.20]. 
(Which is) the very thing it was required to show. 



t In modern notation, this proposition states that if a : b :: c : d then a : b :: a — c : b — d, where all symbols denote numbers. 



'Edv Saiv onoaoiouv dpi'djiol dvdXoyov, earai (be de 
tGv fjyou^ievwv npoc; eva tGv snojisvwv, outmc dnavTee oi 
f)YOU[i£voi npoc dnavrac touc snojisvoue. 



Proposition 12* 

If any multitude whatsoever of numbers are propor- 
tional then as one of the leading (numbers is) to one of 
the following so (the sum of) all of the leading (numbers) 
will be to (the sum of) all of the following. 



A B r A 

'EaT«aav onoaoiouv dpid^ioi dvdXoyov oi A, B, T, A, 
(be; 6 A npoc tov B, out«c 6 T npoc; tov A- Xey«, oti eotiv 
(be; 6 A npoc; tov B, outmc oi A, T npoc; touc B, A. 

'End yap scrav (be; 6 A npoc; tov B, outmc 6 T npoc 
tov A, o apa ^lipoe eotIv 6 A tou B rj \±epf], to ai)TO [lepoc 
sojI xai 6 T tou A rj \xepr]. xal ouva[i(p6Tepoc apa 6 A, 
T auva^itpoTepou tou B, A to auTO [ispoe eaTiv fj Ta ai)Ta 
[ispf], dnep 6 A tou B. eaTiv apa (be 6 A npoc tov B, outwc 
oi A, T npoc touc B, A- onep e5ei 5eic;ai. 



A B C D 

Let any multitude whatsoever of numbers, A, B, C, 
D, be proportional, (such that) as A (is) to B, so C (is) 
to D. I say that as A is to B, so A, C (is) to B, D. 

For since as A is to B, so C (is) to D, thus which(ever) 
part, or parts, A is of £?, C is also the same part, or parts, 
of D [Def. 7.20]. Thus, the sum A, C is also the same 
part, or the same parts, of the sum B, D that A (is) of B 
[Props. 7.5, 7.6]. Thus, as A is to S, so A, C (is) to B, D 
[Def. 7.20]. (Which is) the very thing it was required to 
show. 



205 



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t In modern notation, this proposition states that if a : b :: c : d then a:6::a + c:6 + d, where all symbols denote numbers. 



'Edv Teaaapec; dpidjiol dvdXoyov Saiv, xal evaXXdc^ 
dvdXoyov eaovTai. 



Proposition 13 f 

If four numbers are proportional then they will also 
be proportional alternately. 



A B r A 

"Eaxwaav Teaaapec; apid^ioi dvdXoyov oi A, B, T, A, 
6 A Tipoc; tov B, outoc; 6 T Tipoc; tov A- Xeyw, oti xal 

evaXXai; dvdXoyov eaovxai, (be; 6 A Tipoc; tov T, outcoc; 6 B 

Tipoc; tov A. 

Tkel ydp eaTiv (be; 6 A Tipoc; tov B, outck 6 T Tipoc; tov 
A, o dpa \±spoz eaTiv 6 A tou B fj \±spr\, to auTo \±spoz tov. 
xal 6 T tou A fj Ta ai)Ta [ispt]. evaXXai; dpa, S jiepoe; eaTiv 
6 A tou T fj [iepi], to ai)To ^tepoc; larl xal 6 B tou A fj Ta 
auTa nepT). eaTiv dpa q<;6 A Tipoc; tov T, outw<; 6 B Tipoc; 
tov A- oTiep e8el 8el5ai. 

t In modern notation, this proposition states that if a : b :: c : d then a : 

18'. 

'Eav Saiv onoaoiouv dpi-djiol xal dXXoi auToTe; I'ooi to 
TiXfjiEtoc; auvSuo Xa^pavojisvoL xal ev tQ auTfi Xoycp, xal 5i' 
I'oou ev t(0 auTW Xoyw eaovTai. 

A' 1 A i 1 

B' 1 Ei 1 

' Z' ' 

'EaTwaav oTtoaoiouv dpiiSu-ol oi A, B, T xal dXXoi auToTe; 
Taoi to TtXfj'doc; auvSuo Xa^tpavouevoi ev iu auT« Xoyw oi 
A, E, Z, &>q (iev 6 A Tipoc; tov B, outw<; 6 A Tipoc; tov E, 
(be 5e 6 B Tipoc; tov T, outwc; 6 E Ttp6<; tov Z- Xeyw, oti 
xal 8i° Taou eaTiv d;6A Tipoc; tov T, outwc; 6 A Tipoc; tov 
Z. 

'End ydp eaTiv (be; 6 A Tipoc; tov B, outmc; 6 A Tipoc; 
tov E, evaXXdc; dpa eaTiv (be; 6 A Tipoc; tov A, outwc 6 B 
Tipoc; tov E. ndXiv, inzi eaTiv (be; 6 B npoc; tov T, outwc; 6 



A B C D 

Let the four numbers A, B, C, and D be proportional, 
(such that) as A (is) to B, so C (is) to D. I say that they 
will also be proportional alternately, (such that) as A (is) 
to C, so B (is) to D. 

For since as A is to B, so C (is) to D, thus which(ever) 
part, or parts, A is of B, C is also the same part, 
or the same parts, of D [Def. 7.20]. Thus, alterately 
which (ever) part, or parts, A is of C, B is also the same 
part, or the same parts, of D [Props. 7.9, 7.10]. Thus, as 
A is to C, so B (is) to D [Def. 7.20]. (Which is) the very 
thing it was required to show. 

::b:d, where all symbols denote numbers. 

Proposition 14 f 

If there are any multitude of numbers whatsoever, 
and (some) other (numbers) of equal multitude to them, 
(which are) also in the same ratio taken two by two, then 
they will also be in the same ratio via equality. 

A i ' D 1 

B i 1 E i 1 

Ci 1 F i 1 

Let there be any multitude of numbers whatsoever, A, 
B, C, and (some) other (numbers), D, E, F, of equal 
multitude to them, (which are) in the same ratio taken 
two by two, (such that) as A (is) to B, so D (is) to E, 
and as B (is) to C, so E (is) to F. I say that also, via 
equality as A is to C, so D (is) to F. 

For since as A is to B, so D (is) to E, thus, alternately, 
as A is to D, so B (is) to E [Prop. 7.13]. Again, since 
as B is to C, so E (is) to F, thus, alternately, as B is 



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E npbc, tov Z, evaXXdi; dpa eoxlv cbc; 6 B npbc, tov E, outcoc; to E, so C (is) to F [Prop. 7.13]. And as B (is) to E, 
6 T 7ipo<; tov Z. cbc; 8e 6 B Ttpoc; tov E, outwc; 6 A Ttpoc; so A (is) to D. Thus, also, as A (is) to £>, so C (is) to F. 
tov A- xocl cbc; apa 6 A Ttpoc; tov A, outwc; 6 V npbc, tov Thus, alternately, as A is to C, so L> (is) to F [Prop. 7.13]. 
Z- evaXXdi; apa eaTiv cbc; 6 A Ttpoc; tov V, outw<; 6 A Ttpoc; (Which is) the very thing it was required to show. 
tov Z- onep eSei BeT^oti- 

t In modern notation, this proposition states that if a : b :: d : e and b : c : : e : f then a : c :: d : f, where all symbols denote numbers. 



IS ■ 



Proposition 15 



'Edv [lovac; dpi , f)[i6v Tiva (jtexpf], iaaxic; Be CTepoc; dpid^ioc; 
dXXov Tiva dpidjiov (ieTpfj, xal evaXXdc; iaaxic; f\ ^lovac; tov 
Tphov dprdjiov \±£Tpr]oe\. xal 6 BeuTepoc; tov TETapTov. 



B H 

i 1 1— 



r 

— i 



Ah 



E 

i — 



K 

— i — 



A 

— i — 



Z 

— i 



Movdc; yap f] A dpi%6v Tiva tov Br ^CTperao, iaaxic; 8e 
STspoc; dpnf)^6<; 6 A aXXov Tiva dpidjiov tov EZ ^CTpeiTcy 
Xeyo, oti xal evaXXdi; iaaxic; f] A ^tovdc; tov A dpi-d^ov 
^iSTpeT xal 6 Br tov EZ. 

'End yap iaaxic; f) A ^lovdc; tov Br dpnf)[i6v ^teTpeT xal 6 
A tov EZ, oaai apa eialv ev tw BT [iovd8ec, ToaouToi elm 
xal ev t« EZ dpid^iol i'aoi tG A. Birjpriadio 6 [lev Br eic; tolc, 
ev eauTW [lovdSac; xac, BH, H9, 8r, 6 8e EZ etc; touc; tw A 
laouc; touc; EK, KA, AZ. eaTai 8r) i'aov to TtXTj-doc; ifiv BH, 
H9, 6r tw TtXyydei twv EK, KA, AZ. xal ercel laai eialv ai 
BH, H9, 9T [iovd8ec dXXrjXaic;, eiai 8e xal oi EK, KA, AZ 
dpid^tol i'aoi dXXr]Xoi<;, xai eaTiv laov to TtXfj'doc; t«v BH, 
H9, 9r ^iovd8tov tG TtXfj'dei t«v EK, KA, AZ dpn9[iwv, 
eaTai apa cbc; f] BH jiovdc; Ttpoc; tov EK dpid^ov, outck f) 
H9 jiovdc; Ttpoc; tov KA dpi'dfiov xal f] 9T jiovdc; upoc; tov 
AZ dpid^iov. eaTai apa xal cbc; eic; twv rjyou^ievcov Ttpoc; eva 
iSv CTto^evwv, outwc" obtavTCc; oi fjyou^tevoi Ttpoc; dTtavTac; 
touc; ctio^icvouc;- eaTiv apa cbc; f) BH ^tovdc; Ttpoc; tov EK 
dpid^iov, outmc; 6 Br upoc; tov EZ. i'ar] Be f) BH jiovdc; Tfj 
A (iovdSi, 6 Be EK dpid^oc; iu A dprd^w. eaTiv apa wc; f) 
A [iovdc; upoc; tov A dpnf)[i6v, outwc; 6 Br upoc; tov EZ. 
iadxic; apa f) A [lovac, tov A dpnJ^iov jiCTpeT xai 6 Br tov 
EZ- oTiep e8ei 8eTc;ai. 



If a unit measures some number, and another num- 
ber measures some other number as many times, then, 
also, alternately, the unit will measure the third num- 
ber as many times as the second (number measures) the 
fourth. 



B 

i— 



G 

— i — 



H 

— i — 



C 

— i 



E 

i— 



K 

— i— 



D 

For let a unit A measure some number BC, and let 
another number D measure some other number EE as 
many times. I say that, also, alternately, the unit A also 
measures the number D as many times as BC (measures) 
EF. 

For since the unit A measures the number BC as 
many times as D (measures) EF, thus as many units as 
are in BC, so many numbers are also in EF equal to 
D. Let BC have been divided into its constituent units, 
BC, GH, and HC, and EF into the (divisions) EK, KL, 
and LF, equal to D. So the multitude of (units) BC, 
GH, HC will be equal to the multitude of (divisions) 
EK, KL, LF. And since the units BC, GH, and HC 
are equal to one another, and the numbers EK, KL, and 
LF are also equal to one another, and the multitude of 
the (units) BG, GH, HC is equal to the multitude of the 
numbers EK, KL, LF, thus as the unit BG (is) to the 
number EK, so the unit GH will be to the number KL, 
and the unit HC to the number LF. And thus, as one of 
the leading (numbers is) to one of the following, so (the 
sum of) all of the leading will be to (the sum of) all of 
the following [Prop. 7.12]. Thus, as the unit BG (is) to 
the number EK, so BC (is) to EF. And the unit BG (is) 
equal to the unit A, and the number EK to the number 
D. Thus, as the unit A is to the number D, so BC (is) to 
EF. Thus, the unit A measures the number D as many 
times as BC (measures) EF [Def. 7.20]. (Which is) the 
very thing it was required to show. 



t This proposition is a special case of Prop. 7.9. 



207 



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Edv Buo dpid^iol TtoXXarcXaaidaavxec; dXXr]Xouc; tiolGoi 
xivac;, oi yevojjievoi. zE, aGxcSv Taoi dXXiqXoic; eaovxai. 

A' ' 

B 1 

r> ' 

A' 1 

E' ' 

TEaxcoaav 5uo dpi'd^oi oi A, B, xal 6 [Lev A xov B tioX- 
XaTiXaaidaac; xov T tcoicixo, 6 Se B xov A TtoXXaTiXaaidaac; 
xov A Ttoieixav Xeyco, oxi igoc; eaxlv 6 T xfl A. 

Tkel yap 6 A xov B TtoXXaTiXaaidaac; xov Y TteTtoirjxev, 
6 B dpa xov T jiexpeT xaxd xdc; ev xfi A jiovdBac;. ^expel 8e 
xal f) E ^.ovdc; xov A dpi/djiov xaxd xd<; ev auxfi [lovdBac;- 
iadxu; dpa f\ E fiovac; xov A dpi/d^iov (jiexpeT xal 6 B xov 
T. evaXXdi; dpa iadxic; f] E jiovdc; xov B dpidjiov JiexpeT xal 
6 A xov r. ndXiv, ETcei 6 B xov A noXXaitXaaidaac; xov A 
nenoiy]xev , 6 A dpa xov A JiexpeT xaxd xdc; ev xfi> B jiovdSac;. 
[iexpel 8e xal i\ E fiovac; xov B xaxd xdc; ev auxcp jiovdSac;- 
iadxic; dpa f] E jiovdc; xov B dpid^iov JiexpeT xal 6 A xov A. 
iadxic; 8e f) E [iovdc; xov B dpi-d^ov ejiexpei xal 6 A xov E 
iadxic; dpa 6 A exdxepov xwv Y, A ^expel. iao<; dpa eaxlv 
6 T xw A- oTiep eSei SeTc^ai. 



Proposition 16 1 " 

If two numbers multiplying one another make some 
(numbers) then the (numbers) generated from them will 
be equal to one another. 

A' ' 

B 1 

Ci ' 

D 1 

Ei 1 

Let A and B be two numbers. And let A make C (by) 
multiplying B, and let B make D (by) multiplying A. I 
say that C is equal to D. 

For since A has made C (by) multiplying B, B thus 
measures C according to the units in A [Def. 7.15]. And 
the unit E also measures the number A according to the 
units in it. Thus, the unit E measures the number A as 
many times as B (measures) C. Thus, alternately, the 
unit E measures the number B as many times as A (mea- 
sures) C [Prop. 7.15]. Again, since B has made D (by) 
multiplying A, A thus measures D according to the units 
in B [Def. 7.15]. And the unit E also measures B ac- 
cording to the units in it. Thus, the unit E measures the 
number B as many times as A (measures) D. And the 
unit E was measuring the number B as many times as 
A (measures) C. Thus, A measures each of C and D an 
equal number of times. Thus, C is equal to D. (Which is) 
the very thing it was required to show. 



t In modern notation, this proposition states that ab = ba, where all symbols denote numbers. 



'Edv dpid^oc; Buo dpid^ouc; TtoXXaTiXaaidaac; Ttoifj xivac, 
oi ysvo^ievoi zl, auxGv xov auxov e^ouai Xoyov xolc; iroXXa- 
TtXaaiaadeTaLv. 

A' ' 

b 1 r 1 

A i 1 E' 1 

Z — ' 

Apid^ioc; Y a P ° A Suo dpid^iouc; xouc; B, Y TtoXXa- 
TiXaaidaac; xouc; A, E Ttoieixw Xeyw, oxi eaxlv wc; 6 B Ttpoc; 
xov T, ouxwc; 6 A Ttpoc; xov E. 

Tkel yap 6 A xov B TtoXXaTiXaaidaac; xov A TteTtoirjxev, 
6 B dpa xov A ^expel xaxd xd<; ev iu A ^lovdSac;. JiexpeT 



Proposition 17 f 

If a number multiplying two numbers makes some 
(numbers) then the (numbers) generated from them will 
have the same ratio as the multiplied (numbers) . 
A' ' 

B i C 1 1 

D 1 E i 1 

F i — i 

For let the number A make (the numbers) D and 
E (by) multiplying the two numbers B and C (respec- 
tively). I say that as B is to C, so D (is) to E. 

For since A has made D (by) multiplying B, B thus 
measures D according to the units in A [Def. 7.15]. And 



208 



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ELEMENTS BOOK 7 



Be xoa f\ Z ^iovd<; xov A dpii9[i6v xaxd xdc; ev auxw [lovdSac 
iadxic; apa i] Z jiovag xov A dpidjiov ^.expeT xal 6 B xov A. 
eaxiv apa cb<; f) Z uovdc Ttpoc; xov A dpidjiov, ouxco<; 6 B 
Ttpoc; xov A. Bid xd auxd Br] xal (bt; f] Z ^.ovdc; Ttp6<; xov A 
dpid^iov, ouxwc; 6 T Ttpoc; xov E- xal foz apa 6 B Ttpoc; xov 
A, ouxwc; 6 T Ttpoc; xov E. evaXXd^ apa eaxlv foe, 6 B Ttpoc; 
xov T, ouxcoc; 6 A Ttpoc; xov E- oTtep eSei Bel^ai. 



the unit F also measures the number A according to the 
units in it. Thus, the unit F measures the number A as 
many times as B (measures) D. Thus, as the unit F is 
to the number A, so B (is) to D [Dei. 7.20] . And so, for 
the same (reasons), as the unit F (is) to the number A, 
so C (is) to E. And thus, as B (is) to D, so C (is) to E. 
Thus, alternately, as B is to C, so D (is) to E [Prop. 7.13]. 
(Which is) the very thing it was required to show. 



t In modern notation, this proposition states that if d = a b and e = ac then d : e :: b : c, where all symbols denote numbers. 



IT]'. 

'Edv Buo dpidjiol dpn&jjiov xiva TtoXXaTtXaaidaavxec; 
tiolGoi xivac;, oi yevo^ievoi ec; auxfiv xov auxov l^oum Xoyov 
xolc; TtoXXaTtXaaidaaaiv. 

A' ' 

B 1 

r> 1 

Ai 1 

Ei 1 

Auo yap dpid^ol oi A, B dpi'djiov xiva xov T TtoXXa- 
TtXaaidaavxec; xouc; A, E Ttoieixwaav Xeyio, oxi eaxiv foe, 6 
A Ttpoc xov B, ouxwc; 6 A Ttpoc; xov E. 

'Etcei yap 6 A xov T TtoXXaTtXaaidaac; xov A TtCTtoirjxev, 
xal 6 T apa xov A TtoXXaTtXaaidaac; xov A TtCTto[r)xev. Bid 
xd auxd Br] xal 6 T xov B TtoXXaTtXaaidaac; xov E TteTtoirjxev. 
dpid^ioc; Br] 6 T Buo dpid^iouc; xouc; A, B TtoXXaTtXaaidaac; 
xouc; A, E Tt£Tio(r]X£v. eaxiv apa foe, 6 A Ttpoc; xov B, ouxoc; 
6 A Ttpoc; xov E- OTtep eBei BeT^ou. 



Proposition 18* 

If two numbers multiplying some number make some 
(other numbers) then the (numbers) generated from 
them will have the same ratio as the multiplying (num- 
bers). 

A' ' 

B' 1 

C' ' 

D 1 

Ei 1 

For let the two numbers A and B make (the numbers) 
D and E (respectively, by) multiplying some number C. 
I say that as A is to B, so D (is) to E. 

For since A has made D (by) multiplying C, C has 
thus also made D (by) multiplying A [Prop. 7.16]. So, for 
the same (reasons), C has also made E (by) multiplying 
B. So the number C has made D and E (by) multiplying 
the two numbers A and B (respectively) . Thus, as A is to 
B, so D (is) to E [Prop. 7.17]. (Which is) the very thing 
it was required to show. 



t In modern notation, this propositions states that if a c = d and bc = e then a : b :: d : e, where all symbols denote numbers. 



I'd'. 

'Edv xeaaapec; dpid^iol dvdXoyov Soiv, 6 ex Ttpcjxou xal 
xexdpxou yevojievoc; dpi/duo? i'aoc; eaxai xfi ex Beuxepou xal 
xpixou yevopivw dprd^iSy xal edv 6 ex Ttpwxou xal xexdpxou 
yevojievoc; dpi/duo? laoc, fj xw ex Beuxepou xal xpixou, oi 
xeaaaapec; dpiduol dvdXoyov eaovxai. 

"Eaxioaav xeaaapec; dpid^ol dvdXoyov oi A, B, T, A, 
tlx; 6 A Ttpoc; xov B, ouxwc; 6 T Ttpoc xov A, xal 6 \iev A 
xov A TtoXXaTtXaaidaac; xov E Ttoieixio, 6 Be B xov T TtoX- 
XaTtXaaidaac; xov Z Ttoieixor Xeyto, oxi'i'aoc; eaxlv 6 E xQ Z. 



Proposition 19* 

If four number are proportional then the number cre- 
ated from (multiplying) the first and fourth will be equal 
to the number created from (multiplying) the second and 
third. And if the number created from (multiplying) the 
first and fourth is equal to the (number created) from 
(multiplying) the second and third then the four num- 
bers will be proportional. 

Let A, B, C, and D be four proportional numbers, 
(such that) as A (is) to B, so C (is) to D. And let A make 
E (by) multiplying D, and let B make F (by) multiplying 
C. I say that E is equal to F. 



209 



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ABTAEZH ABCDEFG 



'O yap A xov E TtoXXaTtXaoidoac; xov H Ttoieixco. STtri 
ouv 6 A tov r TtoXXaTtXaoidoac; xov H TtSTtoirjXEv, xov 8e 
A TtoXXaTtXaoidoac; xov E TtSTtoirjxev, dpid^ioc; 8f) 6 A 86o 
dpiiSjiouc; xouc; E, A TtoXXaTtXaoidoac; xouc; H, E TteTtoirjxev. 
eoxiv dpa u<; 6 T Ttpoc; xov A, ouxioc; 6 H Ttpoc; xov E. dXX' 

6 T Ttpoc; xov A, ouxgjc; 6 A Ttpoc; xov B- xal tbc; dpa 
6 A Ttpoc; xov B, ouxcoc; 6 H Ttpoc; xov E. TtdXiv, tm\ 6 A 
xov r TtoXXaTtXaoidoac; xov H TteTtoiiqxsv, dXXd \xr]v xal 6 
B xov T TtoXXaTtXaoidoac; xov Z Tt£Ttoir]X£v, Buo 8/) dpii9|jiol 
oi A, B dpi'djiov xiva xov E TtoXXaTtXaoidoavxsc; xouc; H, Z 
Tt£Ttoir|xaoiv. soxiv dpa d>c; 6 A Ttpoc; xov B, ouxioc; 6 H Ttpoc; 
xov Z. dXXd ^irjv xal wcoA Ttpoc; xov B, ouxgjc; 6 H Ttpoc; 
xov E' xal d>c; dpa 6 H Ttpoc; xov E, ouxioc; 6 H Ttpoc; xov 
Z. 6 H dpa Ttpoc; sxdxspov xSv E, Z xov auxov zyzi Xoyov 
I'ooc; dpa eoxiv 6 E xfi Z. 

'Eoxio 8rj ndXiv I'ooc; 6 E x« Z- Xsyco, oxi saxlv 6c; 6 A 
Ttpoc; xov B, ouxwc; 6 E Ttpoc; xov A. 

Tcbv yap auxcbv xaxaaxeuaoiJevxwv, ind I'ooc; eoxlv 6 
E xa> Z, eoxiv dpa cbc; 6 H Ttpoc; xov E, ouxcoc; 6 H Ttpoc; xov 
Z. dXX' tbc; ^i£v 6 H Ttpoc; xov E, ouxok 6 E Ttpoc; xov A, tbc; 
8e 6 H Ttpoc; xov Z, ouxgjc; 6 A Ttpoc; xov B. xal tbc; dpa 6 A 
Ttpoc; xov B, ouxtoc; 6 E Ttpoc; xov A- onep eBsi 8eTc;ai. 



For let A make G (by) multiplying C. Therefore, since 
A has made G (by) multiplying C, and has made E (by) 
multiplying D, the number A has made G and by mul- 
tiplying the two numbers C and D (respectively). Thus, 
as C is to D, so G (is) to £ [Prop. 7.17]. But, as C (is) to 

D, so A (is) to B. Thus, also, as A (is) to _B, so G (is) to 
i?. Again, since A has made G (by) multiplying G, but, 
in fact, B has also made F (by) multiplying C, the two 
numbers A and £? have made G and F (respectively, by) 
multiplying some number C. Thus, as A is to B, so G (is) 
to F [Prop. 7.18]. But, also, as A (is) to B, so G (is) to 

E. And thus, as G (is) to E, so G (is) to F. Thus, G has 
the same ratio to each of E and F. Thus, E is equal to F 
[Prop. 5.9]. 

So, again, let E be equal to F. I say that as A is to -B, 
so G (is) to £>. 

For, with the same construction, since E is equal to F, 
thus as G is to E, so G (is) to F [Prop. 5.7]. But, as G 
(is) to £, so G (is) to £> [Prop. 7.17]. And as G (is) to F, 
so A (is) to B [Prop. 7.18]. And, thus, as A (is) to B, so 
G (is) to D. (Which is) the very thing it was required to 
show. 



t In modern notation, this proposition reads that if a : b :: c : d then ad = be, and vice versa, where all symbols denote numbers. 



X . 

Oi sXd/ioxoi dpidjjiol twv xov auxov Xoyov £)(6vxtov 
auxolc; [lexpouoi xouc; xov auxov Xoyov e)(ovxa<; iodxic; o 
xe ^d^tov xov ^id^ova xal 6 eXdoowv xov eXdooova. 

'Eoxwoav yap eXd)(ioxoi dpidjioi xfiv xov auxov Xoyov 
EXovxcov xou; A, B oi EA, EZ- Xeyto, oxi iodxic; 6 EA xov 
A ^expEi xal 6 EZ xov B. 



Proposition 20 

The least numbers of those (numbers) having the 
same ratio measure those (numbers) having the same ra- 
tio as them an equal number of times, the greater (mea- 
suring) the greater, and the lesser the lesser. 

For let CD and EF be the least numbers having the 
same ratio as A and B (respectively) . I say that CD mea- 
sures A the same number of times as EF (measures) B. 



210 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



A B 



T r " 


r Et 




© 


H 






z 


A 





A B 



C 

G1 

D 



E 

H-j 
F 



'O TA ydp xou A oux eaxi [iepr). e£ yap 8uvax6v, screw 
xal 6 EZ apa xou B xa auxa [ispf] eaxiv, amp 6 TA xou 
A. oaa apa saxlv ev xio TA y.spt] xou A, xoaauxd eoxi xal 
sv xfi> EZ [ispf] xou B. 8i./]pr]crf>M 6 jji;v TA sic; xa xou A 
\iept] xa TH, HA, 6 Se EZ sic; xa xou B [iepf] xa EO, OZ- 
saxai 8/) taov xo TtXrydoc; xtov TH, HA xG TiAr^si xfiv E6, 
OZ. xal ztzei tool rioiv oi TH, HA dpnSjiol dXXiqXoic, rial 
8e xal oi EO, OZ dpi-Ojioi Taoi dXXr|Xoic;, xai saxiv laov xo 
TtXrji9oc; iwv TH, HA ifi TiX^ei xtov EO, 6Z, eaxiv apa ibc; 
6 TH Ttpoc; xov EO, ouxtoc; 6 HA rcpoc; xov OZ. saxai apa 
xal ok ric; xtov fjyoujjivtov 7tp6<; eva xtov eno^evwv, ouxtoc; 
anavxec; oi f|you^ievoi upoc; anavxag xouc; sttojisvouc;. eaxiv 
apa tog 6 TH Ttpoc; xov EO, ouxtog 6 TA Ttpoc; xov EZ' oi 
TH, EO apa xolc; TA, EZ ev xtp auxto Xoytp eialv eXdaaovec; 
ovxec; auifiv ouep eaxlv dSuvaxov uicoxeivxai yap oi FA, 
EZ eXd^iaxoi xtov xov auxov Xoyov e)(6vxtov auxolc;. oux 
apa |iipr] saxlv 6 TA xou A- jiepoc; apa. xal 6 EZ xou B xo 
auxo ^tepoc; eaxiv, onep 6 TA xou A- iadxic; apa 6 TA xov 
A ^expeT xai 6 EZ xov B- ouep e8ei 8el£ai. 



For CD is not parts of A. For, if possible, let it be 
(parts of A) . Thus, EF is also the same parts of B that 
CD (is) of A [Def. 7.20, Prop. 7.13]. Thus, as many parts 
of A as are in CD, so many parts of B are also in EF. Let 
CD have been divided into the parts of A, CG and CD, 
and EF into the parts of B, EH and HF. So the multi- 
tude of (divisions) CG, GD will be equal to the multitude 
of (divisions) EH, HF. And since the numbers CG and 
GD are equal to one another, and the numbers EH and 
HF are also equal to one another, and the multitude of 
(divisions) CG, GD is equal to the multitude of (divi- 
sions) EH, HF, thus as CG is to EH, so GD (is) to HF. 
Thus, as one of the leading (numbers is) to one of the 
following, so will (the sum of) all of the leading (num- 
bers) be to (the sum of) all of the following [Prop. 7.12]. 
Thus, as CG is to EH, so CD (is) to EF. Thus, CG 
and EH are in the same ratio as CD and EF, being less 
than them. The very thing is impossible. For CD and 
EF were assumed (to be) the least of those (numbers) 
having the same ratio as them. Thus, CD is not parts of 
A. Thus, (it is) a part (of A) [Prop. 7.4]. And EF is the 
same part of B that CD (is) oi A [Def. 7.20, Prop 7.13]. 
Thus, CD measures A the same number of times that EF 
(measures) B. (Which is) the very thing it was required 
to show. 



xa'. 

Oi upCSxoi upog dXXfjXouc; dpid^toi eXd/iaxoi eiai xtov xov 
auxov Xoyov c)(6vxtov auxoTc 

'Eaxtoaav Tcptoxoi Txpoc dXXr|Xou<; apid^ol oi A, B- Xeyto, 
oxi oi A, B eXd)(Laxoi rim xtov xov auxov Xoyov e^ovxtov 
auxolc;. 

E£ yap \xr\, eaovxai xivec; xfiv A, B eXdaaovec; dpidjjiol 
ev xto auxto Xoyto ovxec; xou; A, B. eaxtoaav oi T, A. 



Proposition 21 

Numbers prime to one another are the least of those 
(numbers) having the same ratio as them. 

Let A and B be numbers prime to one another. I say 
that A and B are the least of those (numbers) having the 
same ratio as them. 

For if not then there will be some numbers less than A 
and B which are in the same ratio as A and B. Let them 
be C and D. 



211 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



A B r A E ABCDE 



Tkel ouv oi eXd/ioToi dpidjiol tov tov auxov Xoyov 
£)(6vtov (leTpouai xou<; tov auxov Xoyov e^ovxag iadxic; 
6 xe ^.ei^ov tov jiei^ova xal 6 eXaTTOv tov cXaxxova, 
TouTeaxiv o ts ?]you^evo<; xov f)you^evov xal 6 en6[ie\oc, 
tov eTto^ievov, iadxu; apa 6 T tov A jieTpel xal 6 A tov B. 
oadxic 5r) 6 T tov A jieTpeT, ToaaDTai ^iovd8e<; eaToaav ev 
iw E. xal 6 A apa tov B \xejpel xaxa tote, ev to E ^iovd8ac;. 
xal E7cd 6 r tov A (iCTpel xaTa tuq ev iu E ^tovd8a<;, xat 6 
E apa tov A \Lexpei xaTa Tag ev to T jiovdBac;. 8id Ta auTa 
8r] 6 E xal tov B ^eTpel xaTa Tag ev iS A ^lovdBag. 6 E 
apa tou<; A, B (iCTpel 7tpoTou<; ovTac; Tipog dXXr)Xou<;' OTiep 
sotIv dSuvaTov. oux apa eaovTai Tiveg tov A, B eXdaaove<; 
dpiiL>[iol ev to ai)TO Xoyo ovtcc; toTc A, B. oi A, B apa 
eXd)(ioTo[ eloi tov tov auTov Xoyov e^ovTCov auTolc;- ojtep 
e8ei 8eTc;ai. 



x(3'. 

Oi sXdxiaToi dpi%ol iwv tov auTov Xoyov e^ovTOv 
auTolc; upoToi npbc, dXXf|Xouc; eiaiv. 

A' 1 

B 1 

r 1 

Ai 1 

Ei 1 

TEaToaav eXdxicrcoi dprd^oi tov tov auTov Xoyov 
cXovtov auTou; oi A, B- Xeyo, oti oi A, B TtpoToi 7tp6<; 
dXXrjXoug rialv. 

Ei yap \vf\ eiai TtpoToi izpbz dXXiqXouc;, \LSTpr\os\. uc, 
auToug dprdjioc;. [leTperco, xal eaTO 6 T. xal oadxic [iev 
6 T tov A [jiETpEi, ToaaDTai jiovdBec; eaToaav ev to A, 



Therefore, since the least numbers of those (num- 
bers) having the same ratio measure those (numbers) 
having the same ratio (as them) an equal number of 
times, the greater (measuring) the greater, and the lesser 
the lesser — that is to say, the leading (measuring) the 
leading, and the following the following — C thus mea- 
sures A the same number of times that D (measures) B 
[Prop. 7.20] . So as many times as C measures A, so many 
units let there be in E. Thus, D also measures B accord- 
ing to the units in E. And since C measures A according 
to the units in E, E thus also measures A according to 
the units in C [Prop. 7.16]. So, for the same (reasons), E 
also measures B according to the units in D [Prop. 7.16]. 
Thus, E measures A and B, which are prime to one an- 
other. The very thing is impossible. Thus, there cannot 
be any numbers less than A and B which are in the same 
ratio as A and B. Thus, A and B are the least of those 
(numbers) having the same ratio as them. (Which is) the 
very thing it was required to show. 

Proposition 22 

The least numbers of those (numbers) having the 
same ratio as them are prime to one another. 

A' 1 

Bi 1 

Ci 1 

Di 1 

Ei 1 

Let A and B be the least numbers of those (numbers) 
having the same ratio as them. I say that A and B are 
prime to one another. 

For if they are not prime to one another then some 
number will measure them. Let it (so measure them), 
and let it be C. And as many times as C measures A, so 



212 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



oadxic; 8e 6 T tov B [izxpei, xooauxai ^iovd8ec; eaxwaav ev 
t5 E. 

'ETtri 6 T tov A [iZTpei xaxd tolc, ev ifi A (iovd8ac;, 6 
r dpa tov A TtoXXaTtXaaidaac; tov A TtCTtoiiqxev. Sid toc 
auxd 8r) xal 6 T tov E TtoXXaTtXaaidaac; tov B TtCTtoirjxev. 
dpid^toc; Br] 6 T Suo dpid^touc; touc; A, E TtoXXaTtXaaidaac; 
touc; A, B TtCTtoirjxev eaTiv dpa tlx; 6 A Ttpoc; tov E, outoc; 
6 A Ttpoc; tov B- oi A, E dpa toTc A, B ev t« auTO Xoyw 
siaiv eXdaaovsc; ovtsc; auTfiv oitep sotiv dBuvaTov. oux 
dpa touc; A, B dpidjiouc; dpi-d^oc; tic; [LSTpr]aei. oi A, B dpa 
TtpfiToi Ttpoc; dXXrjXouc; eioiv oTtep sSei 8eic;ai. 

xy'. 

°Edv 80o apidfjiol TtpCnoi Ttpoc; dXXr]Xouc; SSmv, 6 tov eva 
auTGv (jiETpGv apiS^cx; Ttpoc; tov Xoittov TtpfiToc; eaxai. 



A B r A 

'Eaxcoaav 8uo api'djj.oi TtpcSxoi Ttpoc; aXkr\kovc oi A, B, 
xov Se A ^xexpsixo tic dpi%6<; 6 r- Xeyw, oxi xal oi T, B 
Ttp&TOi Ttpoc dXXf]Xou<; daiv. 

Ei yap \ir\ eiaiv oi T, B TtpcoToi Ttpoc; dXXrjXouc;, [iETpiqaei 
[tic] touc r, B dpiduoc;. ^tSTpeiTW, xal sotw 6 A. sitei 6 
A tov T usTpei, 6 8e T tov A ^expei, xal 6 A dpa tov A 
^STpei. [i€Tpei 8s xal tov B- 6 A dpa touc; A, B ^expei 
TtpcVtouc; ovTac; Ttpoc; aXkr\kovc onep eaTiv dSuvaTov. oux 
dpa touc T, B dpi'O^iouc; dpi-duoc; tic; ^exprpei. oi T, B dpa 
TtpCnoi Ttpoc; dXXf]Xouc; eiaiv oitep e8ei 8eTc;ai. 

x5'. 

'Edv Suo dpi'd^ioi Ttpoc; Tiva dpidjiov itpwToi Saiv, xal 6 
zi auT«v yzv6\xzvoc Ttpoc; tov auTov itpwToc; saTai. 



many units let there be in D. And as many times as C 
measures B, so many units let there be in E. 

Since C measures A according to the units in D, C 
has thus made A (by) multiplying D [Def. 7.15]. So, for 
the same (reasons), C has also made B (by) multiplying 
E. So the number C has made A and B (by) multiplying 
the two numbers D and E (respectively). Thus, as D is 
to E, so A (is) to B [Prop. 7.17]. Thus, D and E are in 
the same ratio as A and B, being less than them. The 
very thing is impossible. Thus, some number does not 
measure the numbers A and B. Thus, A and B are prime 
to one another. (Which is) the very thing it was required 
to show. 

Proposition 23 

If two numbers are prime to one another then a num- 
ber measuring one of them will be prime to the remaining 
(one) . 



A B C D 

Let A and B be two numbers (which are) prime to 
one another, and let some number C measure A. I say 
that C and B are also prime to one another. 

For if C and B are not prime to one another then 
[some] number will measure C and B. Let it (so) mea- 
sure (them), and let it be D. Since D measures C, and C 
measures A, D thus also measures A. And (D) also mea- 
sures B. Thus, D measures A and B, which are prime 
to one another. The very thing is impossible. Thus, some 
number does not measure the numbers C and B. Thus, 
C and B are prime to one another. (Which is) the very 
thing it was required to show. 

Proposition 24 

If two numbers are prime to some number then the 
number created from (multiplying) the former (two num- 
bers) will also be prime to the latter (number) . 



213 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



A B r A E Z ABCDEF 



Auo ydp dpid^oi oi A, B upog xiva dpidjiov xov T np&xoi 
eaxtoaav, xcd 6 A xov B iraXXaTtXaaidaac; xov A ttoiclxw 
Xeyw, oxi oi T, A Ttpfixoi Tipog dXXr|Xou<; eiaiv. 

Ei yap (ir| eiaiv oi T, A upSnoi upoc; dXXr|Xouc;, [iexpiqae!. 
[tic;] xoug r, A dpii}|j.6<;. ^expeixw, xod eaxto 6 E. xai enei 
oi T, A TCpcdxoi Tipoc; dXXr|Xou<; eiaiv, xov 8s T jiexpeT xic; 
dpii9(ji6c 6 E, oi A, E dpa npSxoi Ttpoc; (xXkr\kovc, eiaiv. 
bouxiq 8/] 6 E xov A [iexpeT, xoaauxai ^lovdSeg eaxtoaav ev 
xw Z- xai 6 Z dpa xov A (lexpEi xaxa xdg ev xto E ^iovd5a<;. 
6 E dpa xov Z TtoXXaTtXaaidaat; xov A TteTtoirjxev. dXXd 
(iT]v xai 6 A xov B TioXXanXaaidaag xov A Tie7io[r)X£v lao<z 
dpa eaxiv 6 ex xfiv E, Z xfi ex xwv A, B. edv 8s 6 utto 
x£5v dxpwv i'oog fj xa> Otco xcov [leawv, oi xeaaapec; dpii&jjioi 
dvdXoyov eiaiv eaxiv dpa cl>g 6 E Ttpoc; xov A, ouxcoc 6 B 
Tipoc; xov Z. oi 5e A, E upcoxoi, oi 8e npfixoi xai eXdxiaxoi, oi 
8e eXdxiaxoi dpi'djioi xaiv xov auxov Xoyov e)(6vxcov auxou; 
(icxpouai xoug xov auxov Xoyov e^ovxac; iadxic; o xe [ieii^iov 
xov ^ei^ova xai 6 eXdaatov xov eXdaaova, xouxeaxiv 6 xe 
f)you[ievo<; xov fjyoujievov xai 6 CTrajievoc; xov eirajievov 6 
E dpa xov B ^expeT. [icxpeT 8e xai xov T- 6 E dpa xoug B, T 
|iexpei upcoxoui; ovxac rcpoc; dXXrjXouc;- orcep eoxiv d8uvaxov. 
oux dpa xoug r, A dpidjiouc; dpid^toc xi<; [lexp^aei. oi T, A 
dpa Tipfixoi rcpoc; dXXiqXouc; eiaiv onep e8ei SeT^ai. 



For let A and B be two numbers (which are both) 
prime to some number C. And let A make D (by) multi- 
plying B. I say that C and D are prime to one another. 

For if C and D are not prime to one another then 
[some] number will measure C and D. Let it (so) mea- 
sure them, and let it be E. And since C and A are prime 
to one another, and some number E measures C, A and 
E are thus prime to one another [Prop. 7.23]. So as 
many times as E measures D, so many units let there 
be in F. Thus, F also measures D according to the units 
in E [Prop. 7.16]. Thus, E has made D (by) multiply- 
ing F [Def. 7.15]. But, in fact, A has also made D (by) 
multiplying B. Thus, the (number created) from (multi- 
plying) E and F is equal to the (number created) from 
(multiplying) A and B. And if the (rectangle contained) 
by the (two) outermost is equal to the (rectangle con- 
tained) by the middle (two) then the four numbers are 
proportional [Prop. 6.15]. Thus, as E is to A, so B (is) 
to F. And A and E (are) prime (to one another). And 
(numbers) prime (to one another) are also the least (of 
those numbers having the same ratio) [Prop. 7.21]. And 
the least numbers of those (numbers) having the same 
ratio measure those (numbers) having the same ratio as 
them an equal number of times, the greater (measuring) 
the greater, and the lesser the lesser — that is to say, the 
leading (measuring) the leading, and the following the 
following [Prop. 7.20]. Thus, E measures B. And it also 
measures C. Thus, E measures B and C, which are prime 
to one another. The very thing is impossible. Thus, some 
number cannot measure the numbers C and D. Thus, 
C and D are prime to one another. (Which is) the very 
thing it was required to show. 



xe'. 

'Edv 86o dpi%oi TtpGxoi 7tp6<; dXXr|Xou<; waiv, 6 ex xou 
evo<; auxfiv yevojjievoc; Ttpoc; xov Xoitiov TtpGxoc; eaxai. 

TEaxcoaav 8uo dpi%oi Ttpwxoi npoc, dXXr|Xou<; oi A, B, 
xai 6 A eauxov TtoXXaTtXaaidaag xov T noieixw Xey«, oxi 



Proposition 25 

If two numbers are prime to one another then the 
number created from (squaring) one of them will be 
prime to the remaining (number). 

Let A and B be two numbers (which are) prime to 



214 



ETOIXEIfiN C- 



oi B, T itpwxoi icpoc; dXX/jXouc; eiaiv. 

A B r A 



Ksicrdco ydp x£> A Taoc; 6 A. eicei oi A, B icpaixoi icpoc; 
dXXf]Xou<; Eiaiv, laoc, 8e 6 A t« A, xai oi A, B dpa icpGxoi 
icpoc; &XXr|Xou<; eiaiv exdxepoc; dpa xt3v A, A icpoc; xov B 
icpwxoc; eaxiv xal 6 ex xcov A, A dpa yevojievoc; icpoc; xov B 
icpwxoc; eaxai. 6 8e ex xov A, A yevojievoc; apidfjioc; eaxiv 6 
T. oi T, B dpa icpoxoi icpoc; dXXr|Xouc; eiaiv oicep eSei SeTc^ai. 



'Edv 8uo dpid^oi icpoc; 8uo dpid^iouc; d[icpoxepoi icpoc; 
exdxepov icpwxoi Saiv, xai oi eZ, auxwv yevojievoi icpcoxoi 
icpoc; dXXr]XoU(; eaovxai. 

A' 1 r> ' 

B' 1 A i ' 

E' ' 

Z' 1 

Auo yap dpidjioi oi A, B icpoc; 80o dpi'djioug xouc T, 
A d^tcpoxepoi Ttpoc; exdxepov icpGxoi eaxcoaav, xai 6 jiev 
A xov B icoXXaicXaaidaac; xov E icoieixco, 6 8e T xov A 
icoXXaicXaaidaac; xov Z icoieixw Xeyw, oxi oi E, Z icpfixoi 
icpoc; dXXr]Xouc; eiaiv. 

'Etc! yap exdxepoc; xov A, B icpoc xov T icpCSxoc; eaxiv, 
xai 6 ex x«v A, B dpa yevo^ievoc; icpoc; xov T icpwxoc; eaxai. 
6 8e ex xwv A, B yevo^ievoc; eaxiv 6 E - oi E, T dpa icpfixoi 
icpoc; dXXf]Xou<; eiaiv. Sid xa auxa 8r) xai oi E, A icpcoxoi 
icpoc; dXXr]Xouc; eiaiv. exdxepoc; dpa xwv T, A Ttpoc; xov E 
icpwxoc; eaxiv. xai 6 ex xGv T, A dpa yevo^evoc; icpoc; xov 
E icpfixoc; eaxai. 6 8e ex xwv T, A yevo^tevoc; eaxiv 6 Z. oi 
E, Z dpa icpwxoi icpoc; dXXr]Xou<; eiaiv oicep eSei Selc^ai. 



ELEMENTS BOOK 7 



one another. And let A make C (by) multiplying itself. I 
say that B and C are prime to one another. 

A B C D 



For let D be made equal to A. Since A and B are 
prime to one another, and A (is) equal to D, D and B are 
thus also prime to one another. Thus, D and A are each 
prime to B. Thus, the (number) created from (multily- 
ing) D and A will also be prime to B [Prop. 7.24]. And C 
is the number created from (multiplying) D and A. Thus, 
C and B are prime to one another. (Which is) the very 
thing it was required to show. 

Proposition 26 

If two numbers are both prime to each of two numbers 
then the (numbers) created from (multiplying) them will 
also be prime to one another. 

A i ' C' ' 

B i 1 Di 1 

Ei ' 

F ' ' 

For let two numbers, A and B, both be prime to each 
of two numbers, C and D. And let A make E (by) mul- 
tiplying B, and let C make F (by) multiplying D. I say 
that E and F are prime to one another. 

For since A and B are each prime to C, the (num- 
ber) created from (multiplying) A and B will thus also 
be prime to C [Prop. 7.24]. And E is the (number) cre- 
ated from (multiplying) A and B. Thus, E and C are 
prime to one another. So, for the same (reasons), E and 
D are also prime to one another. Thus, C and D are each 
prime to E. Thus, the (number) created from (multiply- 
ing) C and D will also be prime to E [Prop. 7.24]. And 
F is the (number) created from (multiplying) C and D. 
Thus, E and F are prime to one another. (Which is) the 
very thing it was required to show. 



215 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



'Eav Buo dpid^ioi Tipfixoi npog dXXf|Xouc; Saiv, xal tioX- 
XomAaai&aai; exdxepo^ eauTov uoif) xiva, oi yevo^ievoi ec; 
auT<3v Tipoxoi npog dXXr]Xou<; eaovxai, xav oi e<; apxrjg 
tou<; yevo^tevoui; TToXXaTtXaaidaavTec; noi&ai Tivag, xdxeTvoi 
TipwToi Kpo<; dXXr]Xou<; eaovTai [xal del nepl touc, axpouc, 
touto au^paivei]. 

A B T A E Z 



Proposition 27 f 

If two numbers are prime to one another and each 
makes some (number by) multiplying itself then the num- 
bers created from them will be prime to one another, and 
if the original (numbers) make some (more numbers by) 
multiplying the created (numbers) then these will also be 
prime to one another [and this always happens with the 
extremes] . 

A B C D E F 



"EaTwaav Buo dpi%ol Kpcoxoi 7tp6<; dXXr|Xou<; oi A, B, 
xal 6 A eauTov [lev 7ioXXa7iXaaidaa<; tov T itoierao, tov 
Be r 7toXXaTtXaaidaa<; tov A tcoicitco, 6 Be: B eauTov [iev 
TtoXXanXaaidaat; tov E koieltw, tov Be E noXXaTtXaaidaac; 
tov Z tioisitw Xeyto, oti oi' ts T, E xal oi A, Z TtpCrcoi npbc, 
dXXf]Xou<; eiaiv. 

'End yap oi A, B TtpCrcoi npb<z dXXr|Xou<; eiaiv, xal 6 
A eauTov TioXXanXaoidoag tov T TteTto[r)xev, oi T, B apa 
TtpCrcoi Ttpoc; dXXr|Xou<; eiaiv. kizzi ouv oi T, B TtpcoToi Ttpoc; 
dXXr]Xouc; eiaiv, xal 6 B eauTov TtoXXarcXaaidaac; tov E 
TteTtoirjxev, oi T, E apa TtpCkoi Ttpoc; dXXiqXouc; eiaiv. TtdXiv, 
end oi A, B TtpfiToi Ttpoc; dXXr|Xouc; eiaiv, xal 6 B eauTov 
TtoXXaTtXaaidaac; tov E TteTtoirjxev, oi A, E apa jipwToi Ttpoc; 
dXXiqXoug eiaiv. enel ouv Buo dpiOjiol oi A, T Ttpoc; Buo 
dpii9[jioU(; touc; B, E djicpoTepoi Ttpoc; exaTepov TtpCrcoi eiaiv, 
xal 6 ex tC5v A, T apa yevojievoc; Ttpoc; tov ex tGv B, E 
upwToc; eoTiv. xai cotiv 6 |iev ex iSv A, T 6 A, 6 Be ex 
tCSv B, E 6 Z. oi A, Z apa TtpCrcoi Ttpoc; dXXrjXouc; eiaiv 
oTtep eBei BeT^ai. 



Let A and B be two numbers prime to one another, 
and let A make C (by) multiplying itself, and let it make 
D (by) multiplying C. And let B make E (by) multiplying 
itself, and let it make F by multiplying E. I say that C 
and E, and D and F, are prime to one another. 

For since A and B are prime to one another, and A has 
made C (by) multiplying itself, C and B are thus prime 
to one another [Prop. 7.25]. Therefore, since C and B 
are prime to one another, and B has made E (by) mul- 
tiplying itself, C and E are thus prime to one another 
[Prop. 7.25]. Again, since A and B are prime to one an- 
other, and B has made E (by) multiplying itself, A and 
E are thus prime to one another [Prop. 7.25]. Therefore, 
since the two numbers A and C are both prime to each 
of the two numbers B and E, the (number) created from 
(multiplying) A and C is thus prime to the (number cre- 
ated) from (multiplying) B and E [Prop. 7.26] . And D is 
the (number created) from (multiplying) A and C, and F 
the (number created) from (multiplying) B and E. Thus, 
D and F are prime to one another. (Which is) the very 
thing it was required to show. 



t In modern notation, this proposition states that if a is prime to b, then a 2 is also prime to b 2 , as well as a 3 to 6 3 , etc., where all symbols denote 
numbers. 



XT]'. 

'Eav Buo dpi%oi TtpcoToi Ttpoc; dXXiqXouc; Saiv, xal au- 
vajicpoTepoc; Ttpoc; exaTepov auTCSv TtpSTOc; eaTai' xal eav 
auvajicpoTepoc; Ttpoc; eva Tiva auTWv TtpCrcoc; fj, xal oi zE, 
dpxfjc; dpi'd^ioi TtpCrcoi Ttpoc; dXXr|Xou<; eaovTai. 



Proposition 28 

If two numbers are prime to one another then their 
sum will also be prime to each of them. And if the sum 
(of two numbers) is prime to any one of them then the 
original numbers will also be prime to one another. 



216 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



A Br 

i 1 1 

A' 1 

Euyxeicrdoaav yap 860 dpi/d^ioi Txpoxoi npbc, dXXiqX- 
ouc ol AB, BP Xeyw, oxi xal auva^tcpoxepoc 6 Ar Ttpoc 
exdxepov xfiv AB, Br Ttpwxoc saxiv. 

El yap \ir\ sicuv ol TA, AB Ttpwxoi Ttpoc dXXr]Xouc, 
^t£Tpf]a£i xic xouc TA, AB dpidjioc. ^.expeixw, xal saxw 6 A. 
ETte! ouv 6 A xouc TA, AB ^isxpeT, xal Xomov dpa xov Br 
^Exprpa. [lezpei Be xal xov BA- 6 A dpa xouc AB, Br ^ie- 
xpel icpcoxouc ovxac icpoc dXXrjXouc 6%ep eaxlv dBuvaxov. 
oux dpa xouc FA, AB dpidjiouc dpid^ioc xic ^exp^aei- oi 
TA, AB dpa icpoxoi Ttpoc dXXr|Xouc eia'w. 8id xa auxd Br| 
xal oi Ar, TB Ttpwxoi npbc, dXXf|Xouc eiaiv. 6 TA dpa Ttpoc 
exdxepov xfiv AB, Br TtpCSxoc eaxiv. 

"Eaxcoaav Br] TtdXiv oi FA, AB Ttpcoxoi Ttpoc dXXr)Xouc 
Xeyw, 6x1 xal oi AB, Br Ttpfixoi Ttpoc dXXr]Xouc eiaiv. 

EE yap (Jir) eiaiv oi AB, Br Ttpwxoi npbc, dXXr]Xouc, 
[Lejpfiaei xic xouc AB, Br dpid^toc. ^lexpeixw, xal eaxw 6 A. 
xal £7i si 6 A exdxepov xwv AB, Br ^texpeT, xal oXov dpa xov 
TA (iexpr]oei. ^lexpeT Be xal xov AB- 6 A dpa xouc FA, AB 
^.expel Ttpwxouc ovxac Ttpoc dXXf|Xouc" oTtep eaxiv dBuvaxov. 
oux dpa xouc AB, Br dpidjiouc dpnf)|i6c tic; ^texpr]aei. oi 
AB, Br dpa Ttpoxoi npbc, dXXr]Xouc eiaiv OTtep eBei SeT^ai. 



"Auac npCSxoc dpnJ^oc Ttpoc ditavxa dpii^ov, ov \j.t) \±e- 
xpeT, Kpwxoc eaxiv. 

A' ' 

Bi 1 

ri 1 

'Eaxw Ttpwxoc dprd^toc 6 A xal xov B [!/] jiexpeixGr Xeyio, 
oxi oi B, A TtpCkoi Ttpoc dXXrjXouc eiaiv. 

Ei yap ^ir] eiaiv oi B, A Ttpaixoi Ttpoc dXXrjXouc, ^texpf|aei 
tic auxouc dpid^ioc. (iexpeixw 6 T. inei 6 T xov B [icxpel, 
6 Be A xov B ou ^expeT, 6 T dpa xw A oux eaxiv 6 auxoc. 
xal ETiei 6 T xouc B, A ^.expeT, xal xov A dpa (lexpel TtpGxov 
ovxa [ir] ov auxfi 6 auxoc OTtep eaxiv dBuvaxov. oux dpa 
xouc B, A jiexprjoei xic dpid^ioc. oi A, B dpa upwxoi npbc, 
dXXf]Xouc eioiv OTtep eBei SeT^ai. 



A B C 

1 1 1 

Di 1 

For let the two numbers, AB and BC, (which are) 
prime to one another, be laid down together. I say that 
their sum AC is also prime to each of AB and BC. 

For if CA and AB are not prime to one another then 
some number will measure CA and AB. Let it (so) mea- 
sure (them), and let it be D. Therefore, since D measures 
CA and AB, it will thus also measure the remainder BC. 
And it also measures BA. Thus, D measures AB and 
BC, which are prime to one another. The very thing is 
impossible. Thus, some number cannot measure (both) 
the numbers CA and AB. Thus, CA and AB are prime 
to one another. So, for the same (reasons), AC and CB 
are also prime to one another. Thus, CA is prime to each 
of AB and BC. 

So, again, let CA and AB be prime to one another. I 
say that AB and BC are also prime to one another. 

For if AB and BC are not prime to one another then 
some number will measure AB and BC. Let it (so) mea- 
sure (them), and let it be D. And since D measures each 
of AB and BC, it will thus also measure the whole of 
CA. And it also measures AB. Thus, D measures CA 
and AB, which are prime to one another. The very thing 
is impossible. Thus, some number cannot measure (both) 
the numbers AB and BC. Thus, AB and BC are prime 
to one another. (Which is) the very thing it was required 
to show. 

Proposition 29 

Every prime number is prime to every number which 
it does not measure. 

A' 1 

B' 1 

C' ' 

Let A be a prime number, and let it not measure B. I 
say that B and A are prime to one another. For if B and 
A are not prime to one another then some number will 
measure them. Let C measure (them) . Since C measures 
B, and A does not measure B, C is thus not the same as 
A. And since C measures B and A, it thus also measures 
A, which is prime, (despite) not being the same as it. 
The very thing is impossible. Thus, some number cannot 
measure (both) B and A. Thus, A and B are prime to 
one another. (Which is) the very thing it was required to 



217 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



show. 



X'. 

'Edv 860 dpid^iol TtoXXauXaaidaavTec; dXXr|Xou<; TtoiCkri 
xiva, tov 8e ysvojisvov z\ auxCSv ^lexprj tic; Tipfixoc; dpi%6c;, 
xal sva xGv z\ «PX*K (jieTpiqoeL. 

A' 1 

B 1 

r> 1 

Ai 1 

Ei 1 

Auo yap dpi-djjiol oi A, B noXXaitXaaidaavTec; dXXr|Xouc; 
xov T Ttoisraoaav, tov 8e T ^leTpeiTW tic; TtpCkoc; dpi%6c; 6 
A - Xfyco, oti 6 A sva iSv A, B [lETpsI. 

Tov yap A \±r\ (iSTpsiTW xai soti up«Toc; 6 A- oi A, 
A apa 7ip«Toi Ttpoc; dXXr|Xouc; riaiv. xal oadxig 6 A tov T 
(iSTpsI, ToaaOTai [iovdSec; eoT«oav sv tco E. end ouv 6 A 
tov r [LsxpeX xoltol Tag ev tG E jiovdBac;, 6 A apa tov E 
TioXXaTiXaaidoac; tov T TtSTtoi/jxev. dXXa [iy]\i xal 6 A tov 
B TioXXaTiXaoidaac; tov T Tieuoirjxsv I'ooc; apa sotIv 6 £x 
twv A, E tu ex twv A, B. eaTiv apa 6 A Ttpoc; tov A, 
outwc; 6 B Ttpoc; tov E. oi Be A, A rcpfinoi, oi 5e upwToi xal 
eXd)(ioToi, oi 8e eXd)(iaToi ^iSTpouai touc; tov auTov Xoyov 
EXovTac; iadxic; 6 ts ^idCtov tov ^.ei^ova xal 6 eXdoawv tov 
eXdaaova, toutsotiv o te fjyoupievoc; tov fjyoupievov xal 6 
STio^ievoc; tov enojievov 6 A apa tov B (leTpsi. ojioicoc; Br| 
8dc;o^ev, oti xal iav tov B \ir] (jLexpfj, tov A ^STprjaa. 6 A 
apa eva iSv A, B ^isTpeT- oitsp sSei Bel^ai. 



Proposition 30 

If two numbers make some (number by) multiplying 
one another, and some prime number measures the num- 
ber (so) created from them, then it will also measure one 
of the original (numbers). 

A' 1 

Bi 1 

Ci 1 

Di 1 

Ei 1 

For let two numbers A and B make C (by) multiplying 
one another, and let some prime number D measure C. I 
say that D measures one of A and B. 

For let it not measure A. And since D is prime, A 
and D are thus prime to one another [Prop. 7.29]. And 
as many times as D measures C, so many units let there 
be in E. Therefore, since D measures C according to 
the units E, D has thus made C (by) multiplying E 
[Def. 7.15]. But, in fact, A has also made C (by) multi- 
plying B. Thus, the (number created) from (multiplying) 
D and E is equal to the (number created) from (mul- 
tiplying) A and B. Thus, as D is to A, so B (is) to E 
[Prop. 7.19]. And D and A (are) prime (to one another), 
and (numbers) prime (to one another are) also the least 
(of those numbers having the same ratio) [Prop. 7.21], 
and the least (numbers) measure those (numbers) hav- 
ing the same ratio (as them) an equal number of times, 
the greater (measuring) the greater, and the lesser the 
lesser — that is to say, the leading (measuring) the lead- 
ing, and the following the following [Prop. 7.20]. Thus, 
D measures B. So, similarly, we can also show that if 
(D) does not measure B then it will measure A. Thus, 
D measures one of A and B. (Which is) the very thing it 
was required to show. 



Xa'. 

"Auac; auvdevToc; dpid[i6c; Otto npwTou tivoc; dpi'dpiou ^s- 
TpeiTai. 

'Ecnxo auvdsvToc; dpi%6<; 6 A- Xeyw, oti 6 A (mo TipwTou 
tivoc; dpii9(jio0 (jiETpslTai. 

Tkel yap auvdsToc; eaTiv 6 A, \LZTpr\az\. tic; auTov 



Proposition 31 

Every composite number is measured by some prime 
number. 

Let A be a composite number. I say that A is measured 
by some prime number. 

For since A is composite, some number will measure 
it. Let it (so) measure (A), and let it be B. And if B 



218 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



6Lpv&\i6c,. ^ETpeiTW, xal eaxw 6 B. xal si \±ev np&xdz saxiv 6 
B, yEyovoc; av sTr) xo £7uxax$£v. el Be auvdexcx;, [i£xpf|a£i 
tic; auxov dpid^ioc;. [lExpdxw, xal eaxw 6 T. xal sitsl 6 T 
xov B [iexpsT, 6 8s: B xov A [iexpsT, xal 6 T apa xov A 
[isxpeT. xal si (lev Ttpfixoc; eaxiv 6 T, yeyovoc; av sir) xo 
ETiixax'Osv. ei 8s auvdexoc;, [iexpf|aei xi? auxov dpi%6<;. 
xoiauxrjc; 8r) yivo[ievr)<; eiuaxeijiewc; X/jcp-drjaexai xic; TtpGxoc; 
dpid^toc;, oc; [iexpf|aei. ei yap ou X/jcp-drjoexai, [lexprpouai 
xov A dpid^iov omeipoi dpid^toi, Sv exepoc exepou eXdaatov 
eaxiv oitep eaxiv dSuvaxov ev dpnJ^olc;. Xrjcp'driaexai xic; apa 
Ttpwxoc; dpi'd^xot;, oc; [lexprpei xov Tipo eauxou, oc; xal xov A 
^Expfjaei. 

A' ■ 

B 1 

r> 1 

"Auac; apa auvdevxoc; dpid[x6c; utio Ttpcoxou xivoc; dpid^iou 
[icxpelxai- orcep e5ei SeTc;ai. 

X(3'. 

"Aitac; dpid^toc; fjxoi TtpGxoc; eaxiv fj utio upwxou xivoc; 
dprd^ioO [xexpelxai. 

' A' 1 

TCaxco dpi%6<; 6 A- Xeyto, oxi 6 A f]xoi Ttpcoxoc; eaxiv fj 
utio Tipwxou xivoc; dpid^tou [icxpelxai. 

El [iev ouv Ttpwxoc; eaxiv 6 A, yeyovoc; av €ir\ xo 
CTiixax'Oev. ei Be auvdexoc;, [iexpr|aei xu; auxov TtpGxoc; 
dpid \±6c,. 

"Arcac; apa dpiiiJjjioc; rjxoi TtpGxoc; eaxiv fj utio upwxou 
xivoc; dpn5[jio0 [xexpelxai- onep e8ei SeT^ai. 

Xy'. 

Api-djjiov Bo'devxwv ouoawvouv eupelv xouc; eXaxiaxouc; 
xwv xov auxov Xoyov exovxcov auxolc;. 

"Eaxwaav ol SoiSevxec; orcoaoiouv dpid^iol ol A, B, T' 8eT 
8r) eupelv xouc; eXaxiaxouc; xwv xov auxov Xoyov exovxcov 
xou; A, B, T. 

Ol A, B, r yap fjxoi upfixoi Ttpoc; dXXfjXouc; eialv fj ou. 
ei [i£v ouv ol A, B, T Tipoxoi rcp6<; dXXrjXouc; claw, eXdxiaxoi 
eiai xwv xov auxov Xoyov exovxov auxolc;. 



is prime then that which was prescribed has happened. 
And if (B is) composite then some number will measure 
it. Let it (so) measure (B), and let it be C. And since 
C measures B, and B measures A, C thus also measures 
A. And if C is prime then that which was prescribed has 
happened. And if (C is) composite then some number 
will measure it. So, in this manner of continued inves- 
tigation, some prime number will be found which will 
measure (the number preceding it, which will also mea- 
sure A) . And if (such a number) cannot be found then an 
infinite (series of) numbers, each of which is less than the 
preceding, will measure the number A. The very thing is 
impossible for numbers. Thus, some prime number will 
(eventually) be found which will measure the (number) 
preceding it, which will also measure A. 

A' 1 

B 1 

O 1 

Thus, every composite number is measured by some 
prime number. (Which is) the very thing it was required 
to show. 

Proposition 32 

Every number is either prime or is measured by some 
prime number. 

A i 1 

Let A be a number. I say that A is either prime or is 
measured by some prime number. 

In fact, if A is prime then that which was prescribed 
has happened. And if (it is) composite then some prime 
number will measure it [Prop. 7.31]. 

Thus, every number is either prime or is measured 
by some prime number. (Which is) the very thing it was 
required to show. 

Proposition 33 

To find the least of those (numbers) having the same 
ratio as any given multitude of numbers. 

Let A, B, and C be any given multitude of numbers. 
So it is required to find the least of those (numbers) hav- 
ing the same ratio as A, B, and C. 

For A, B, and C are either prime to one another, or 
not. In fact, if A, B, and C are prime to one another then 
they are the least of those (numbers) having the same 
ratio as them [Prop. 7.22]. 



219 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



ABTAEZHOKAM 

1 1 1 I I 1 1 1 1 1 I 



Et 8e ou, eiXiqcp'do xov A, B, T to ^eyiaxov xoivov 
^iixpov 6 A, xai oadxic; 6 A exaaxov xov A, B, T jiexpe'i, 
xoaauxai ^.ovd8e<; eaxoaav ev exdaxo xov E, Z, H. xai 
exaaxoc apa xov E, Z, H exaaxov xov A, B, T [icxpeT xaxa 
xa<; ev xo A jiovd8a<;. oi E, Z, H apa xou<; A, B, T ladxi? 
^texpouaiv oi E, Z, H apa xoic; A, B, T ev xo auxo Xoyo 
eiaiv. Xeyo 8rj, oxi xai eXd)(iaxoi. ei yap \ifi eiaiv oi E, Z, 
H eXd)(iaxoi xov xov auxov Xoyov e)(6vxov xoi<; A, B, T, 
eaovxai [xivec] xov E, Z, H eXdaaovec; dpid^ioi ev tu auxo 
Xoyo ovxec xou; A, B, T. eaxoaav oi 0, K, A- iadxu; apa 6 
xov A jiexpel xai exdxepo<; xov K, A exdxepov iwv B, T. 
oadxu; Se 6 xov A ^expeT, xoaauxai ^iovd8e<; eaxoaav ev 
xo M- xai exdxepo<; apa xov K, A exdxepov xov B, T ^texpei 
xaxa xa<; ev xo M ^iovd8a<;. xai eitei 6 xov A ^.expeT xaxa 
xac; ev iu M ^iovd8a<;, xai 6 M apa xov A jiexpei xaxa xa<; 
ev xo [iovdSa<;. 8ia xa auxa Sr| 6 M xai exdxepov xov B, 
r jiexpei xaxa xac; ev exaxepo xov K, A ^iovd8ac 6 M apa 
xou<; A, B, r ^.expeT. xai eitei 6 xov A jiexpei xaxa xa<; 
ev xo M jiovdSac;, 6 apa xov M 7ioXXan;Xaaidaa<; xov A 
7ieTio(r]xev. 8ia xa auxa 8r) xai 6 E xov A TtoXXaTtXaaidaac; 
xov A 7ieTio(r]xev. iao<; apa eaxiv 6 ex xov E, A xo ex 
iwv 0, M. eaxiv apa (i><; 6 E 7tp6<; xov 0, ouxo<; 6 M 7tp6<; 
xov A. ^isi^tov Se 6 E xou ©• ^eii^ov apa xai 6 M xou A. 
xai ^xexpei xoug A, B, E oitep eaxiv dBuvaxov tmoxeixai 
yap 6 A xov A, B, T xo (jieyiaxov xoivov (icxpov. oux apa 
eaovxai xive<; xov E, Z, H eXdaaove<; dpud^oi ev xo auxo 
Xoyo ovxec; xou; A, B, T. oi E, Z, H apa eXd)(iaxo[ eiai xov 
xov auxov Xoyov exovxov xoic A, B, E oiiep e8ei Sel^ai. 



X8'. 

Auo dpid^ov Bo'devxov eupeiv, ov eXd)(iaxov jiexpouaiv 
dprd^iov. 

'Eaxwaav oi BoiSevxec; 8uo dpnSjioi oi A, B- Sei Sr) eupeiv, 



ABCDEFGHKLM 

1 1 I 1 1 I 1 1 I 1 I 



And if not, let the greatest common measure, D, of 

A, B, and C have be taken [Prop. 7.3]. And as many 
times as D measures A, B, C, so many units let there 
be in E, F, G, respectively And thus E, F, G mea- 
sure A, B, C, respectively, according to the units in D 
[Prop. 7.15]. Thus, E, F, G measure A, B, C (respec- 
tively) an equal number of times. Thus, E, F, G are in 
the same ratio as A, B, C (respectively) [Def. 7.20]. So I 
say that (they are) also the least (of those numbers hav- 
ing the same ratio as A, B, C). For if E, F, G are not 
the least of those (numbers) having the same ratio as A, 

B, C (respectively), then there will be [some] numbers 
less than E, F, G which are in the same ratio as A, B, C 
(respectively). Let them be H, K, L. Thus, H measures 
A the same number of times that K, L also measure B, 

C, respectively. And as many times as H measures A, so 
many units let there be in M. Thus, K, L measure B, 
C, respectively, according to the units in M. And since 
H measures A according to the units in M, M thus also 
measures A according to the units in H [Prop. 7.15]. So, 
for the same (reasons), M also measures B, C accord- 
ing to the units in K, L, respectively. Thus, M measures 
A, B, and C. And since H measures A according to the 
units in M, H has thus made A (by) multiplying M. So, 
for the same (reasons), E has also made A (by) multiply- 
ing D. Thus, the (number created) from (multiplying) 
E and D is equal to the (number created) from (multi- 
plying) H and M. Thus, as E (is) to H, so M (is) to 
D [Prop. 7.19]. And E (is) greater than H. Thus, M 
(is) also greater than D [Prop. 5.13]. And (M) measures 
A, B, and C. The very thing is impossible. For D was 
assumed (to be) the greatest common measure of A, B, 
and C. Thus, there cannot be any numbers less than E, 
F, G which are in the same ratio as A, B, C (respec- 
tively). Thus, E, F, G are the least of (those numbers) 
having the same ratio as A, B, C (respectively) . (Which 
is) the very thing it was required to show 

Proposition 34 

To find the least number which two given numbers 
(both) measure. 

Let A and B be the two given numbers. So it is re- 



220 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



ov eXd)(iaTov ^texpouaiv dpi'd^ov. 

A i 1 B' 1 

ri 1 

Ai 1 

Ei ' Z ' 

Oi A, B yap f\ioi rcpcoxoi Ttpoc; dXXfjXouc; eiaiv fj ou. 
eaxcoaav rcpoxepov oi A, B Tipwxoi rcpoc; dXXrjXouc;, xai 6 A 
xov B TxoXXanXaaidaac; xov T Ttoieixw xdi 6 B apa xov A 
TioXXomXaaidaoK; xov T 7i£K0ir]xev. oi A, B apoc xov T \is- 
xpouaiv. Xeyco 8r], Sxi xai sXd)(iaxov. ei yap \±r\, \±£Tpr\aouoi 
xiva dpii^ov oi A, B iXdaaova ovxa xou T. ^.expeixwaav 
xov A. xai oadxu; 6 A xov A ^xexpeT, xoaauxai jiovd8e<; 
eaxwaav ev x£> E, oadxu; 8e 6 B xov A ^expert, xoaauxai 
|jovd8ec; eaxioaav ev xfi Z. 6 [lev A apa xov E noXka- 
TtXaaidaac; xov A TteTrairjxev, 6 8e B xov Z TroXXaTiXaaidaac; 
xov A TCSTtoi/jxev i'aoc; apa eaxiv 6 ex xcov A, E xai ex xwv 
B, Z. eaxiv apa cl>g 6 A 7tp6<; xov B, ouxgjc; 6 Z npb<z xov 
E. oi 8e A, B Ttpfixoi, oi 8e Ttpfixoi xai eXd)(iaxoi, oi Se 
eXd/iaxoi [icxpouai xouc; xov auxov Xoyov e^ovxac; iadxic; o 
xe (j.£L^cov xov [iciCova xai 6 eXdaawv xov eXdaaova- 6 B 
apa xov E jiexpe'i, tb<; £it6[i£voc eno^ievov. xai snel 6 A xouc; 
B, E KoXXajiXaoidoac; xouc; T, A Ttenoi/jxev, eaxiv apa tbc; 
6 B Ttpoc; xov E, ouxgjc; 6 T npbq xov A. [lexpe'i 8e 6 B xov 
E- piexpeT apa xai 6 T xov A 6 ^ei^cov xov eXdaaova- onep 
eaxiv d8uvaxov. oux apa oi A, B jiexpouai xiva dpidjiov 
eXdaaova ovxa xou T. 6 T apa eXd/iaxoc; &v Otto xfiv A, B 
^expsixai. 



A i 1 B i 1 

Z' 1 Ei ' 

r i 1 

A' 1 

Hi 1 ©i 1 

Mr) eaxwaav 8f] oi A, B Ttpfixoi Ttpoc; dXXrjXouc;, 
xai siXricp'dwaav eXd^iaxoi dpi-djioi x«v xov auxov Xoyov 
EXovxcov xou; A, B oi Z, E- Xaoc, apa eaxiv 6 ex xwv A, E iw 



quired to find the least number which they (both) mea- 
sure. 

A i 1 B i 1 

C' 1 

Di 1 

E i 1 F i 1 

For A and B are either prime to one another, or not. 
Let them, first of all, be prime to one another. And let A 
make C (by) multiplying B. Thus, B has also made C 
(by) multiplying A [Prop. 7.16]. Thus, A and B (both) 
measure C. So I say that (C) is also the least (num- 
ber which they both measure) . For if not, A and B will 
(both) measure some (other) number which is less than 
C. Let them (both) measure D (which is less than C). 
And as many times as A measures D, so many units let 
there be in E. And as many times as B measures D, 
so many units let there be in F. Thus, A has made D 
(by) multiplying E, and B has made D (by) multiply- 
ing F. Thus, the (number created) from (multiplying) 
A and E is equal to the (number created) from (multi- 
plying) B and F. Thus, as A (is) to B, so F (is) to E 
[Prop. 7.19]. And A and B are prime (to one another), 
and prime (numbers) are the least (of those numbers 
having the same ratio) [Prop. 7.21], and the least (num- 
bers) measure those (numbers) having the same ratio (as 
them) an equal number of times, the greater (measuring) 
the greater, and the lesser the lesser [Prop. 7.20]. Thus, 
B measures E, as the following (number measuring) the 
following. And since A has made C and D (by) multi- 
plying B and E (respectively), thus as B is to E, so C 
(is) to D [Prop. 7.17]. And B measures E. Thus, C also 
measures D, the greater (measuring) the lesser. The very 
thing is impossible. Thus, A and B do not (both) mea- 
sure some number which is less than C. Thus, C is the 
least (number) which is measured by (both) A and B. 

A i 1 B i 1 

F i 1 E i 1 

C i 1 

D i 1 

G i 1 H i 1 

So let A and B be not prime to one another. And 
let the least numbers, F and E, have been taken having 
the same ratio as A and B (respectively) [Prop. 7.33]. 



221 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



ex tc5v B, Z. xal 6 A tov E TroXXaTtXaaidaac; tov T tioieltw 
xal 6 B apa xov Z TTOXXaTtXaaidaac; tov T tcttoi/jxev oi A, 
B apa tov T ^ETpouaiv. Xeyio 8r), oxi xal eXd/iaTov. si yap 
[ir|, (ieTprjoouoi Tiva dpii9|ji6v oi A, B sXdaaova ovTa tou 
r. [iETpsiTcooav tov A. xal oodxig [Lev 6 A tov A [isxpeX, 
ToaauTai ^ovd8e<; eaTCoaav ev t« H, oodxig Se 6 B tov A 
(iSTpsT, ToaauTai jiovdSec; saTwaav ev xfi 8. 6 [lev A apa 
tov H TraXXaTtXaaidaac; tov A TteTronqxsv, 6 8s B tov 9 
7ioXXa7iXaaidoa<; tov A 7T£7TO[r)xev. laoc; apa eotIv 6 ex twv 
A, H tw ex twv B, 0' sgtiv apa 6 A npoc; tov B, outcoc; 
6 9 Ttpoc; tov H. u; 8e 6 A 7tpo<; tov B, outioc; 6 Z npog 
tov E- xal &>c, apa 6 Z Ttpoc; tov E, outgjc; 6 9 upog tov 
H. oi Bs; Z, E sXd/ioToi, oi 8e sXd/ioToi jiSTpouai tou<; tov 
auTov Xoyov e^ovTac; iadxic; o te jiei^mv tov jisi^ova xal 6 
sXdaawv tov sXdaaova- 6 E apa tov H [iSTpel. xal end 6 
A touc; E, H TraXXaTtXaaidaac; touc; T, A kstioi/jxev, ecmv 
apa &>c, 6 E Ttpoc; tov H, outgjc; 6 T Ttpoc; tov A. 6 Be E tov 
H [iSTpel' xal 6 T apa tov A ^ETpsI 6 [isi^cov tov sXdaaova- 
OTtep saw dSuvaTov. oux apa oi A, B [iSTprpouai Tiva 
dpi-djiov eXdaaova ovTa tou T. 6 T apa sXd/iaToc; d>v utto 
tGv A, B [jieTpslTai- OTtep enei 5eic;ai. 



Thus, the (number created) from (multiplying) A and E 
is equal to the (number created) from (multiplying) B 
and F [Prop. 7.19]. And let A make C (by) multiplying 
E. Thus, B has also made C (by) multiplying F. Thus, 
A and B (both) measure C. So I say that (C) is also the 
least (number which they both measure) . For if not, A 
and B will (both) measure some number which is less 
than C. Let them (both) measure D (which is less than 
C) . And as many times as A measures D, so many units 
let there be in G. And as many times as B measures D, 
so many units let there be in H. Thus, A has made D 
(by) multiplying G, and B has made D (by) multiplying 
H. Thus, the (number created) from (multiplying) A and 
G is equal to the (number created) from (multiplying) B 
and H. Thus, as A is to B, so H (is) to G [Prop. 7.19]. 
And as A (is) to B, so F (is) to E. Thus, also, as F (is) 
to E, so (is) to G. And F and are the least (num- 
bers having the same ratio as A and B), and the least 
(numbers) measure those (numbers) having the same ra- 
tio an equal number of times, the greater (measuring) 
the greater, and the lesser the lesser [Prop. 7.20]. Thus, 
E measures G. And since A has made C and D (by) mul- 
tiplying E and G (respectively), thus as E is to G, so C 
(is) to D [Prop. 7.17]. And E measures G. Thus, C also 
measures D, the greater (measuring) the lesser. The very 
thing is impossible. Thus, A and B do not (both) mea- 
sure some (number) which is less than C. Thus, C (is) 
the least (number) which is measured by (both) A and 
B. (Which is) the very thing it was required to show. 



Xe'. 

'Edv 6uo dpi'd^.oi dpid^iov Tiva jiSTpGaiv, xal 6 eXd/iarac; 
un auTGv (jieTpou^ievoc; tov auTov jiSTprpsi. 

A i 1 Bi 1 

r z a 

I 1 1 

Ei 1 

Auo yap dpi%ol oi A, B dpi%6v Tiva tov FA ^e- 
TpeiToaav, eXd)(iaTov 8e tov E- Xeyw, oti xal 6 E tov TA 
(iETpeT. 

Ei yap ou [Lsjpei 6 E tov PA, 6 E tov AZ ^tETpwv 
Xsitctco eauTou sXdaaova tov TZ. xal ETtri oi A, B tov E 
^tSTpoOaiv, 6 8e E tov AZ \izxpei, xal oi A, B apa tov 
AZ ^i£Tpr]aouaiv. ^ETpouai 8e xal 6Xov tov TA- xal XoiTtov 
apa tov rZ [iETprjaouaiv eXdaaova ovTa tou E' oitsp sgtIv 
dSuvaTov. oux apa ou (iSTpsT 6 E tov FA- (JETpsT apa- onep 
ebei 8eic;ai. 



Proposition 35 

If two numbers (both) measure some number then the 
least (number) measured by them will also measure the 
same (number). 

A i 1 B ' 1 

C F D 

i 1 1 

Ei 1 

For let two numbers, A and B, (both) measure some 
number CD, and (let) E (be the) least (number mea- 
sured by both A and B) . I say that E also measures CD. 

For if E does not measure CD then let E leave CF 
less than itself (in) measuring DF. And since A and B 
(both) measure E, and E measures DF, A and B will 
thus also measure DF. And (A and B) also measure the 
whole of CD. Thus, they will also measure the remainder 
CF, which is less than E. The very thing is impossible. 
Thus, E cannot not measure CD. Thus, (E) measures 



222 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



W. 

Tpifiv dpi%Gv 8oi3evx«v supeiv, ov eXd)(iaxov [jie- 
xpouaiv aptduov. 

TSaxwaav oi BoiDevxEi; xpeu; dpid^ol ol A, B, E 8ei 8r) 
eOpsTv, ov eXd)(iaxov ^sxpouaiv dpid^iov. 

A' 1 

B i 1 

r i 1 

a i ' 

E' 1 

Z' ' 

EiXr ! ](p , d« yap Otto 860 xwv A, B eXd)(iaxoc; ^lexpou^ievoc; 
6 A. 6 8f) r xov A rjxoi ^.sxpei f\ ou jisxpei. [lexpsixw 
jcpoxepov. [isxpouoi 8e xai 01 A, B xov A' oi A, B, T 
apa xov A jiexpouaiv. Xeyw S^, oxi xal eXd)(iaxov. el yap 
[ir], [iexpr]aouaiv [xiva] dpid^iov oi A, B, T erXdaaova ovxa 
xou A. ^expeixwaav xov E. inel oi A, B, T xov E ^.expouaiv, 
xal oi A, B apa xov E ^xexpouaiv. xal 6 sXd/iaxot; apa 6ti:6 
xfiv A, B piexpou^evoc; [xov E] jjiexprpei. eXd)(iaxoc; 8e Otto 
xGv A, B |iexpou^ev6<; eaxiv 6 A- 6 A apa xov E |iexpf)aei 
6 (lei^tov xov sXdaaova' ousp eaxiv dSuvaxov. oux apa oi 
A, B, T (lexpiqoouoi xiva dpi%6v sXdaaova ovxa xou A- oi 
A, B, T apa sXd/iaxov xov A jisxpouaiv. 

Mr) ^lExpeixco Sf| TidXiv 6 T xov A, xai eiXiqcp'dio bub xwv 
T, A sXd/iaxoc; ^expou^evoc; dpi%6c; 6 E. end oi A, B 
xov A ^expouaiv, 6 8e A xov E ^expsi, xai oi A, B apa 
xov E [isxpouaiv. jiexpsi 8e xai 6 T [xov E- xai] oi A, B, 
T apa xov E jisxpouaiv. Xeyw 8rj, oxi xal sXd/iaxov. si 
yap [if], |iexpf|aoua[ xiva oi A, B, T eXdooova ovxa xou E. 
(lexpsixwaav xov Z. sicd oi A, B, T xov Z ^xexpouaiv, xai oi 
A, B apa xov Z [lexpouaiv xai 6 eXdxioxog apa bub xwv 
A, B [isxpoujisvog xov Z jiexprjoei. £Xd)(iaxo<; 8e bub xwv 
A, B ^.expou^evoc; eaxiv 6 A' 6 A apa xov Z jiexpe'i. ^.expeT 
8e xal 6 T xov Z - oi A, T apa xov Z ^.expouaiv fiaie xai 6 
eXd)(iaxoc; Otto xwv A, T ^iexpou^ievo<; xov Z jiexpr|aei. 6 8e 
eXd)(iaxo<; (mo xwv T, A ^expou^ievo<; eaxiv 6 E- 6 E apa 
xov Z ^.expeT 6 [icii^wv xov eXdaaova- ouep eoxiv dSuvaxov. 
oux apa oi A, B, T ^expr|aoua[ xiva dpid^tov eXdaaova ovxa 
xou E. 6 E apa eXa^iajoc, wv utio x£Sv A, B, T ^lexpelxai- 
oTiep e8ei SeT^ai. 



(CD). (Which is) the very thing it was required to show. 

Proposition 36 

To find the least number which three given numbers 
(all) measure. 

Let A, B, and C be the three given numbers. So it is 
required to find the least number which they (all) mea- 
sure. 

A 1 1 

B ' ' 

C ' ' 

D 1 1 

E ' 1 

F ' 1 

For let the least (number), D, measured by the two 
(numbers) A and B have been taken [Prop. 7.34]. So C 
either measures, or does not measure, D. Let it, first of 
all, measure (D). And A and B also measure D. Thus, 
A, B, and C (all) measure D. So I say that (D is) also 
the least (number measured by A, B, and C) . For if not, 
A, B, and C will (all) measure [some] number which 
is less than D. Let them measure E (which is less than 
D). Since A, B, and C (all) measure E then A and B 
thus also measure E. Thus, the least (number) measured 
by A and B will also measure [E] [Prop. 7.35]. And D 
is the least (number) measured by A and B. Thus, D 
will measure E, the greater (measuring) the lesser. The 
very thing is impossible. Thus, A, B, and C cannot (all) 
measure some number which is less than D. Thus, A, B, 
and C (all) measure the least (number) D. 

So, again, let C not measure D. And let the least 
number, E, measured by C and D have been taken 
[Prop. 7.34]. Since A and B measure D, and D measures 
E, A and B thus also measure E. And C also measures 
[E] . Thus, A, B, and C [also] measure E. So I say that 
(E is) also the least (number measured by A, B, and C) . 
For if not, A, B, and C will (all) measure some (number) 
which is less than E. Let them measure F (which is less 
than E). Since A, B, and C (all) measure F, A and B 
thus also measure F. Thus, the least (number) measured 
by A and B will also measure F [Prop. 7.35]. And D 
is the least (number) measured by A and B. Thus, D 
measures F. And C also measures F. Thus, D and C 
(both) measure F. Hence, the least (number) measured 
by D and C will also measure F [Prop. 7.35]. And E 



223 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



'Edv dpi/duo? (mo xivog dpidjiou [lexprjxai, 6 [isxpoujisvog 
6(itovu^ov ^£po<; e^si xw fisxpoOvxi. 

A' 1 

B 1 

r> 1 

A' — ' 

Apid^xot; yap 6 A tmo xivo<; apiduoO xoO B ^STpeCcrdco - 
Xsy", oxi 6 A o^iwvu^iov ^xepo<; sxsi xw B. 

'Oadxu; yap 6 B xov A [lexpeT, xoaauxai ^.ovd8e;<; eoxw- 
aav ev xw I\ STtri 6 B xov A [Lexpei xaxd xdc; ev xfi T 
[LovabaQ, ^lexpel Be xal f) A ^tovdc; xov T dpid^iov xaxd xdc; 
ev auxC) ^ovdSag, ladxic; dpa f\ A jiovdc; xov T dpi'djiov ^.e- 
xpeT xal 6 B xov A. evaXXdc; dpa ladxic; f) A \io\>aq xov B 
dpid^tov ^texpcT xal 6 T xov A- o dpa ^tepoc; eaxlv f) A jiovdc; 
xoO B dpid^oO, xo auxo ^icpoc; eaxl xal 6 T xou A. r) 8e A 
jiovdc; xou B dpid^ou \±epoq eaxlv 6^wvu[iov aux£y xal 6 T 
dpa xou A ^tepoc; eaxlv o^xovu^ov xO B. waxe 6 A ^tepoc; 
exei xov T o^ovu^tov ovxa iw B- oitep e8ei 8eTc;ai. 



XT]'. 

'Eav dpi-duoc; ^tepoc; e^Tl oxiouv, Otto o^covu^iou dpi%oO 
piexpr^riaexai xw uepei. 

1 

B 1 

V 1 

A> — i 

Apid^oc; yap 6 A ^.epoc; exexco oxiouv xov B, xal tw B 
[jiepei o^xwvu^oc; eoxw [dpidjioc;] 6 T- Xeyoj, oxi 6 T xov A 
^expel. 

Tkel yap 6 B xou A \±epoq eaxlv o^wvujiov x£> T, eaxi 
8e xal f] A jiovdc; xou T [izpoz o^xovu^ov auxo, o dpa ^iepo<; 



is the least (number) measured by C and D. Thus, E 
measures F, the greater (measuring) the lesser. The very 
thing is impossible. Thus, A, B, and C cannot measure 
some number which is less than E. Thus, E (is) the least 
(number) which is measured by A, B, and C. (Which is) 
the very thing it was required to show. 

Proposition 37 

If a number is measured by some number then the 
(number) measured will have a part called the same as 
the measuring (number). 

A' 1 

B' 1 

O 1 

D' 1 

For let the number A be measured by some number 
B. I say that A has a part called the same as B. 

For as many times as B measures A, so many units 
let there be in C. Since B measures A according to the 
units in C, and the unit D also measures C according 
to the units in it, the unit D thus measures the number 
C as many times as B (measures) A. Thus, alternately, 
the unit D measures the number B as many times as C 
(measures) A [Prop. 7.15]. Thus, which(ever) part the 
unit D is of the number B, C is also the same part of A. 
And the unit D is a part of the number B called the same 
as it (i.e., a £>th part). Thus, C is also a part of A called 
the same as B (i.e., C is the £?th part of A). Hence, A has 
a part C which is called the same as B (i.e., A has a Bth 
part) . (Which is) the very thing it was required to show. 

Proposition 38 

If a number has any part whatever then it will be mea- 
sured by a number called the same as the part. 

A i 1 

Bi 1 

Ci 1 

Di 1 

For let the number A have any part whatever, B. And 
let the [number] C be called the same as the part B (i.e., 
B is the Cth part of A) . I say that C measures A. 

For since B is a part of A called the same as C, and 
the unit D is also a part of C called the same as it (i.e., 



224 



ETOIXEIfiN C- 



ELEMENTS BOOK 7 



saxlv f] A ^lovac xoO T dpid^iou, to auxo [iepo<; Sail xal 6 B 
xoO A- tadxu; apa rj A ^.ova<; xov T dpid^iov ^.expeT xal 6 B 
xov A. evaXXd^ apa iadxu; f] A ^tovac; xov B dpid^iov ^expei 
xal 6 T xov A. 6 T apa xov A ^xexpsi- oTtep eSei Bd^ai. 



Aft'. 

Apuduov eupefv, o<; eXd)(iaxo<; wv e^ei xa Scydevxa ^tcpr]. 

a b r 

i — i i 1 i 1 



D is the Cth part of C), thus which (ever) part the unit D 
is of the number C, B is also the same part of A. Thus, 
the unit D measures the number C as many times as B 
(measures) A. Thus, alternately, the unit D measures the 
number B as many times as C (measures) A [Prop. 7.15] . 
Thus, C measures A. (Which is) the very thing it was 
required to show. 

Proposition 39 
To find the least number that will have given parts. 

ABC 

i — i i 1 i 1 



A E 

i 1 i 



■ D ■ , 



H 



H 



Tiaxco xa So'dsvxa y.epf] xa A, B, T- 8d Br] apid^ov 
supeiv, 6<z eXd)(iaxoc; <J>v s^ei xa A, B, T jiepr). 

"Eaxwaav yap xou; A, B, T ^ispeaiv o^covu^ioi dpid^tol 
oi A, E, Z, xal dXrjcp'dw (mo xGv A, E, Z 6Xd)(iaxo<; \±s- 
xpoujievoc; dprdjioi; 6 H. 

c O H apa o^tcovupia [ispf] £X £l xo ^ A, E, Z. xou; Be A, 
E, Z o^covu^ia [ispr) eaxl xa A, B, T- 6 H apa ex £l T & A, B, 
T (ieprj. Xsyw 8rj, oxi xal eXd)(iaxo<; wv, eE yap ^V), eaxai xu; 
xoO H eXdaowv dpnf)uo<;, o<; e^a xa A, B, T [ispr\. eaxw 6 
6. etieI 6 6 ex £l ™ A, B, T fiepr], 6 6 apa bub o^covu^cov 
dpi-d^cov ^t£xpr]Tf)r]0£xai xou; A, B, T [ispeaiv. xou; Be A, B, 
r \iepeow b\i(x>wy.oi dpn^ot daiv oi A, E, Z- 6 9 apa utco 
xwv A, E, Z jiexpdxai. xa[ eaxiv eXdaawv xou H- orcsp 
eaxlv dBuvaxov. oux apa eaxai xu; xou H iXdaatov dpudu6<;, 
be, e^si xa A, B, T [iepf]- bnep eBei BsT^ai. 



Let A, B, and C be the given parts. So it is required 
to find the least number which will have the parts A, B, 
and C (i.e., an Ath part, a Bth part, and a Cth part). 

For let D, E, and F be numbers having the same 
names as the parts A, B, and C (respectively). And let 
the least number, G, measured by D, E, and F, have 
been taken [Prop. 7.36]. 

Thus, G has parts called the same as D, E, and F 
[Prop. 7.37]. And A, B, and C are parts called the same 
as D, E, and F (respectively). Thus, G has the parts A, 
B, and C. So I say that (G) is also the least (number 
having the parts A, B, and C). For if not, there will be 
some number less than G which will have the parts A, 

B, and C. Let it be H . Since H has the parts A, B, and 

C, H will thus be measured by numbers called the same 
as the parts A, B, and C [Prop. 7.38]. And D, E, and 
F are numbers called the same as the parts A, B, and C 
(respectively). Thus, H is measured by D, E, and F. And 
(H) is less than G. The very thing is impossible. Thus, 
there cannot be some number less than G which will have 
the parts A, B, and C. (Which is) the very thing it was 
required to show. 



225 



226 



ELEMENTS BOOK 8 

Continued Proportion 



tThe propositions contained in Books 7-9 are generally attributed to the school of Pythagoras. 



227 



STOIXEIQN rf. 



ELEMENTS BOOK 8 



a . 

'Eav Saiv oaoiSrjuoxouv apid^ol l^fjc; dvdXoyov, oi 8e 
dxpoi auxfiv Ttpfixoi Tip6<; dXXrjXout; waiv, eXd)(iaxoi eiai 
x£Sv xov auxov Xoyov sxovxcov auxolc 

' Ei — ' 

B ' ' Z ' 

T ' ' H' ' 

A i 1 ®i 1 

TCaxcoaav oTtoaoiouv dpi-d^ol e^r|<; dvdXoyov oi A, B, 
r, A, oi Be dxpoi auxaiv oi A, A, Ttpaixoi Ttpoc; aXkr\kovc, 
eaxcoaav Xeyio, oxi oi A, B, T, A eXd)(iaxoi eiai xfiv tov 
auxov Xoyov e/ovxiov auxolc;. 

Ei yap [jltj, eaxwaav eXdxxovec; xaiv A, B, T, A oi E, 
Z, H, ev xw auxcS Xoyio ovxec; auxolc;. xal end oi A, 
B, T, A ev xtp auxcp Xoycp etal xolc; E, Z, H, 0, xod eaxiv 
i'oov xo KkrjdoQ [x«v A, B, T, A] x£> TiXri'dei [x«v E, Z, H, 
9], 8i' i'oou dpa eaxlv <i>c; 6 A upog xov A, 6 E Ttpoc; xov 
9. oi Be A, A ixpwxoi, oi Bs itpcoxoi xal eXd)(iaxoi, oi 6e 
eXd/iaxoi dpidjiol ^.expouai xouc; xov auxov Xoyov e^ovxac; 
iadxic; o xe (iei^cov xov (jiei^ova xai 6 eXdaawv xov eXdaaova, 
xouxeaxiv o xe fjyou^ievoc xov fjyou^ievov xal 6 en6[ievoc, 
xov E7i6|jievov. jiexpel dpa 6 A xov E 6 [iei^cov xov eXdaaova' 
oTtep eaxiv dSuwxov. oux dpa oi E, Z, H, 9 eXdaaovec; 
ovxec; xwv A, B, T, A sv x£5 auxw X6ya> eiaiv auxolc;. oi A, 
B, r, A dpa eXd)(iaxoi eiai xfiv xov auxov Xoyov e/ovxiov 
auxolc;- oTtep eSei Belial. 



P'- 

Aptduouc; eupeiv ec;rjc; dvdXoyov eXa)([axouc;, oaouc; dv 
eraxd^r) xic;, ev iu SoiSevxi Xoyw. 

TEaxco 6 Bo'delc; Xoyoc; ev eXd)(iaxoic; dpn&iio'ic; 6 xou 
A rcpoc; xov B' Bel 8r) dpid^ouc; eupeiv ec;rjc; dvdXoyov 
eXa/iaxouc;, oaouc; dv xic imia£,r\, sv x« xou A Ttpoc; xov B 
Xoycp. 

'Eiuxexdx'dwaav 8rj xeaaapec;, xal 6 A eauxov uoXXa- 
TtXaaidaac; xov T Tioieixw, xov Be B TioXXauXaaidaac; xov A 
Ttoieixw, xal exi 6 B eauxov noXXaTiXaaidaac; xov E ttoieixgj, 
xal exi 6 A xouc; T, A, E TtoXXaitXaaidaac; xouc; Z, H, 9 
Tioieixw, 6 Be B xov E TtoXXaTtXaaidaac; xov K Ttoieixw. 



Proposition 1 

If there are any multitude whatsoever of continuously 
proportional numbers, and the outermost of them are 
prime to one another, then the (numbers) are the least 
of those (numbers) having the same ratio as them. 

A i ' E i 1 

B i 1 F i 1 

C i 1 G' 1 

D ' H' ' 

Let A, B, C, D be any multitude whatsoever of con- 
tinuously proportional numbers. And let the outermost 
of them, A and D, be prime to one another. I say that 
A, B, C, D are the least of those (numbers) having the 
same ratio as them. 

For if not, let E, F, G, H be less than A, B, C, D 
(respectively), being in the same ratio as them. And since 
A, B, C, D are in the same ratio as E, F, G, H, and the 
multitude [of A, B, C, D] is equal to the multitude [of E, 
F, G, H], thus, via equality as A is to D, (so) E (is) to H 
[Prop. 7.14]. And A and D (are) prime (to one another). 
And prime (numbers are) also the least of those (numbers 
having the same ratio as them) [Prop. 7.21]. And the 
least numbers measure those (numbers) having the same 
ratio (as them) an equal number of times, the greater 
(measuring) the greater, and the lesser the lesser — that 
is to say, the leading (measuring) the leading, and the 
following the following [Prop. 7.20]. Thus, A measures 
E, the greater (measuring) the lesser. The very thing is 
impossible. Thus, E, F, G, H, being less than A, B, C, 
D, are not in the same ratio as them. Thus, A, B, C, D 
are the least of those (numbers) having the same ratio as 
them. (Which is) the very thing it was required to show. 

Proposition 2 

To find the least numbers, as many as may be pre- 
scribed, (which are) continuously proportional in a given 
ratio. 

Let the given ratio, (expressed) in the least numbers, 
be that of A to B. So it is required to find the least num- 
bers, as many as may be prescribed, (which are) in the 
ratio of A to B. 

Let four (numbers) have been prescribed. And let A 
make C (by) multiplying itself, and let it make D (by) 
multiplying B. And, further, let B make E (by) multiply- 
ing itself. And, further, let A make F, G, H (by) mul- 
tiplying C, D, E. And let B make K (by) multiplying 
E. 



228 



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ELEMENTS BOOK 8 



a i 1 r i 1 

B i 1 A i 1 

Ei 1 

Z' ' 

Hi ' 

©i 1 

K 1 

Kal tnel 6 A eauxov [lev TtoXXaTtXaaidaae xov T 
7i£iio(r]X£v, tov 8e B TtoXXaTtXaaidaae xov A kstioitjxev, 
eaxiv apa u<;oA Ttpoe xov B, [ouxcoe] 6 T npbc, xov A. 
TidXiv, ercel 6 [lev A xov B TtoXXauXaaidaae xov A TteTtoi/jxev, 
6 Se B eauxov TtoXXaitXaaidaae xov E 7i£no(r]xev, exdxepoe 
apa tuv A, B xov B rcoXXanXaaidaae exdxepov x(5v A, E 
7i£iio(r]xev. eaxiv apa (be 6 A Ttpoe xov B, ouxwe 6 A Ttpoe 
xov E. dXX'' (be 6 A Ttpoe xov B, 6 T Ttpoe xov A- xal (be 
apa 6 T Ttpoe xov A, 6 A Ttpoe xov E. xal CTtel 6 A xoue T, 
A TtoXXaTtXaaidaae xoue Z, H TteTtoirjxev, eaxiv apa (be 6 T 
Ttpoe xov A, [ouxw<;] 6 Z Ttpoe xov H. (be 8e 6 T Ttpoe xov 
A, ouxwe rjv 6 A Ttpoe xov B- xal (be apa 6 A Ttpoe xov B, 6 
Z Ttpoe xov H. TtdXiv, CTtel 6 A xouc A, E TtoXXaTtXaaidaae 
xouc H, TteTtoirjxev, eaxiv apa (be 6 A Ttpoe xov E, 6 H 
Ttpoe xov 9. dXX' (be 6 A Ttpoe xov E, 6 A Ttpoe xov B. xal 
(be apa 6 A Ttpoe xov B, ouxwe 6 H Ttpoe xov 0. xal excel 
oi A, B xov E TtoXXaitXaaidaavxee xouc 9, K TteTtoifjxaaiv, 
eaxiv apa (be 6 A Ttpoe xov B, ouxwe 6 9 Ttpoe xov K. dXX'' 
(be 6 A Ttpoe xov B, ouxwe 6 xe Z Ttpoe xov H xal 6 H Ttpoe 
xov 9. xal (be apa 6 Z Ttpoe xov H, ouxwe o xe H Ttpoe 
xov 9 xal 6 9 Ttpoe xov K- oi T, A, E apa xal oi Z, H, 
9, K dvdXoyov eiaiv ev x£> xou A Ttpoe xov B Xoyw. Xeyw 
8r), 6xi xal eXd)(iaxoi. eitel yap oi A, B eXd)(iaxo( eiai xwv 
xov auxov Xoyov e)(6vxwv auxole, oi Be eXd)(iaxoi xwv xov 
auxov Xoyov £)(6vxwv Ttpwxoi Ttpoe dXXf|Xoue eiaiv, oi A, B 
apa upwxoi Ttpoe dXXrjXouc eiaiv. xal exdxepoe \±kv xwv A, 
B eauxov TtoXXaTtXaaidaae exdxepov xwv T, E TtCTtoir)xev, 
exdxepov 8e x£>v T, E TtoXXaTtXaaidaae exdxepov xwv Z, K 
Tteito(r]xev oi T, E apa xal oi Z, K TtpGxoi Ttpoe dXXr|Xoue 
eiaiv. eav Se waiv oTtoaoioOv dpid^oi e£/je dvdXoyov, oi 
8e axpoi auxCSv Ttpwxoi Ttpoe dXXr]Xoue Saiv, eXd)(iaxo[ eiai 
xwv xov auxov Xoyov exovxov auxole- oi T, A, E apa xal 
oi Z, H, 9, K eXdxiaxoi eiai xwv xov auxov Xoyov exovxeov 
xou; A, B- oTtep eSei Bel^ai. 



A i 1 O 1 

B i 1 D 1 

Ei 1 

F i 1 

Gi 1 

Hi 1 

Ki 1 

And since A has made G (by) multiplying itself, and 
has made D (by) multiplying B, thus as A is to B, [so] G 
(is) to D [Prop. 7.17]. Again, since A has made D (by) 
multiplying B, and B has made E (by) multiplying itself, 

A, B have thus made D, E, respectively, (by) multiplying 

B, Thus, as A is to B, so D (is) to E [Prop. 7.18]. But, as 
A (is) to B, (so) C (is) to D. And thus as G (is) to D, (so) 
D (is) to E. And since A has made G (by) multiplying 

C, D, thus as C is to L>, [so] F (is) to G [Prop. 7.17]. 
And as G (is) to D, so A was to B. And thus as A (is) 
to B, (so) F (is) to G. Again, since A has made G, 
(by) multiplying I?, E, thus as £> is to E, (so) G (is) to 
H [Prop. 7.17]. But, as D (is) to £, (so) A (is) to B. 
And thus as A (is) to B, so G (is) to H. And since A, B 
have made AT (by) multiplying E, thus as A is to B, 
so H (is) to AT. But, as A (is) to B, so F (is) to G, and 
G to B. And thus as F (is) to G, so G (is) to B, and H 
to A". Thus, G, D, E and B, G, B, AT are (both continu- 
ously) proportional in the ratio of A to B. So I say that 
(they are) also the least (sets of numbers continuously 
proportional in that ratio). For since A and B are the 
least of those (numbers) having the same ratio as them, 
and the least of those (numbers) having the same ratio 
are prime to one another [Prop. 7.22], A and B are thus 
prime to one another. And A, B have made G, B, respec- 
tively, (by) multiplying themselves, and have made F, K 
by multiplying G, B, respectively. Thus, G, E and F, K 
are prime to one another [Prop. 7.27]. And if there are 
any multitude whatsoever of continuously proportional 
numbers, and the outermost of them are prime to one 
another, then the (numbers) are the least of those (num- 
bers) having the same ratio as them [Prop. 8.1]. Thus, G, 

D, E and F, G, H, K are the least of those (continuously 
proportional sets of numbers) having the same ratio as A 
and B. (Which is) the very thing it was required to show. 



229 



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ELEMENTS BOOK 8 



H6pio\j.a. 

'Ex Br) xouxou cpavepov, oxi edv xpelc; dpi/d^ioi zZ,ffc 
dvdXoyov eXd/iaxoi Goi xfiv xov auxov Xoyov e/ovxiov 
auxolg, oi axpov auxfiv xexpdytovoi eiaiv, edv 8e xeaaapec, 
xu[3oi. 

y'- 

'Edv fiaiv otioooiouv apid^ol ec;r]<; dvdXoyov eXd)(iax- 
01 xfiv xov auxov Xoyov C)(6vxmv auxolc;, oi dxpoi auxtov 
Tipwxoi upoc dXXrjXouc; eiaiv. 



A 
B 

r 

A 

A 
M 
N 



IO 



'Eaxtoaav oitoaoiouv dpid^toi E^fj? dvdXoyov eXd/iaxoi 
xwv xov auxov Xoyov e)(6vx«v auxolg oi A, B, T, A- Xeyio, 
6xi oi dxpoi auxwv oi A, A TtpGxoi npbc, dXXrjXouc; eiaiv. 

EiXf](p , dwaav yap 8uo ^iev apidjiol eXd)(iaxoi ev xw xCSv 
A, B, T, A Xoyw oi E, Z, xpsT? 6s oi H, 0, K, xod s^fjc; 
svi TtXeiout;, ecoc; xo Xa|ipav6|i£vov TtXrydoc; 'iaov yevrjxai xw 
TtXyydei xwv A, B, T, A. eiXf|Cp , d«aav xal eaxwaav oi A, M, 
N,H. 

Kai etiei oi E, Z sXd)(iaxoi eiai xwv xov auxov Xoyov 
e)(ovxa>v auxou;, itpcoxoi npbz dXXrjXouc; eiaiv. xal Excel 
exdxepov iSv E, Z eauxov ^.ev TioXXaitXaaidaac; exdxepov 
xuv H, K TtC7ioir)xev, exdxepov 8e x£Sv H, K uoXXa- 
TiXaaidaac; exdxepov xwv A, H KCTioirjxev, xai oi H, K dpa 
xai oi A, H npCSxoi npbq dXXr|Xou<; eiaiv. xai enel oi A, B, 
T, A eXd)(iaxoi eiai xwv xov auxov Xoyov exovxwv auxou;, 
eiai Be xai oi A, M, N, S eXd^iaxoi ev xw auxcS Xoyw ovxec 
xdtc A, B, T, A, xai eaxiv laov xo TtXrydoc; xwv A, B, T, 
A x« TiXr) , dei xwv A, M, N, S, exaaxo<; dpa xwv A, B, T, 
A exdaxw xCSv A, M, N, 5 i'ao<; eaxiv i'ao<; dpa eaxiv 6 
\±ev A x« A, 6 8e A xw S. xai eiaiv oi A, H upwxoi 7ipo<; 
dXXf]Xou<;. xai oi A, A dpa TtpGxoi npoc, dXXrjXouc; eiaiv 
oTiep eBei SeTc;ai. 



Corollary 

So it is clear, from this, that if three continuously pro- 
portional numbers are the least of those (numbers) hav- 
ing the same ratio as them then the outermost of them 
are square, and, if four (numbers), cube. 

Proposition 3 

If there are any multitude whatsoever of continu- 
ously proportional numbers (which are) the least of those 
(numbers) having the same ratio as them then the outer- 
most of them are prime to one another. 
A' 1 Ei 1 G' 1 



B 1 F ' 

—i 

D 

L 
M 
N 
O 

Let A, B, C, D be any multitude whatsoever of con- 
tinuously proportional numbers (which are) the least of 
those (numbers) having the same ratio as them. I say 
that the outermost of them, A and D, are prime to one 
another. 

For let the two least (numbers) E, F (which are) 
in the same ratio as A, B, C, D have been taken 
[Prop. 7.33]. And the three (least numbers) G, H, K 
[Prop. 8.2]. And (so on), successively increasing by one, 
until the multitude of (numbers) taken is made equal to 
the multitude of A, B, C, D. Let them have been taken, 
and let them be L, M, N, O. 

And since E and F are the least of those (numbers) 
having the same ratio as them they are prime to one an- 
other [Prop. 7.22] . And since E, F have made G, K, re- 
spectively, (by) multiplying themselves [Prop. 8.2 corn], 
and have made L, O (by) multiplying G, K, respec- 
tively, G, K and L, O are thus also prime to one another 
[Prop. 7.27]. And since A, B, C, D are the least of those 
(numbers) having the same ratio as them, and L, M, N, 
O are also the least (of those numbers having the same 
ratio as them), being in the same ratio as A, B, C, D, and 
the multitude of A, B, C, D is equal to the multitude of 
L, M, N, O, thus A, B, C, D are equal to L, M, N, O, 
respectively. Thus, A is equal to L, and D to O. And L 
and O are prime to one another. Thus, A and D are also 
prime to one another. (Which is) the very thing it was 



230 



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ELEMENTS BOOK 8 



5'. 

Aoywv SoiSevxov oTioatovouv ev eXayJaxoic; dpid^iou; 
dpi%ou<; eupelv e^fjc; dvdXoyov eXayJaxouc; ev xoT<; Bo'deTai 



Xoyou;. 

' B 

r i 1 a 

E' ' Z 

N 1 ' © 

Hi ' H 

M' ' K 

O' ' A 



"Eaxwaav oi SoiSevxec Xoyoi ev eXa)(iaxoic; dpi-d^iou; 6 
xe xoO A Tipoc; xov B xal 6 xou T Tip6<; xov A xod exi 6 
xoO E npoc, xov Z- 8eT 5r) dpid^iouc; eupelv e^rjc; dvdXoyov 
eXayJaxouc; ev xe iS xou A Tipoc; xov B Xoytp xal ev xw xou 
r npoc xov A xal exi iu xou E Tipoc; xov Z. 

EiXVjcp'dcd yap 6 utio xov B, T eXdyiaxoc; [iexpou^evo<; 
dpi%6<; 6 H. xal oadxi? y.ev 6 B xov H ^xexpeT, xoaauxdxic 
xal 6 A xov 6 ^texpeixw, oadxn; 8e 6 T xov H ^texpel, xo- 
aauxdxic xal 6 A xov K jiexpeixw. 6 Se E xov K fjxoi ^texpel 
f\ ou jiexpel. ^texpelxw Tipoxepov. xal oadxu; 6 E xov K fie- 
xpeT, xoaauxdxic xal 6 Z xov A piexpeixo. xal CTtei iadxic; 6 
A xov 9 ^texpeT xal 6 B xov H, eaxiv apa ox 6 A Tipoc; xov 
B, ouxw<; 6 9 Tipoc xov H. Sid xa auxa Srj xal ob<; 6 T Tipoc; 
xov A, ouxgk 6 H Ttpoc xov K, xal exi foe, 6 E Tipoc; xov Z, 
ouxoc; 6 K Tipoc xov A- oi 9, H, K, A apa e^fjc dvdXoyov 
eiaiv ev xe xw xou A Tipoc xov B xal ev iu xou T Tipoc xov 
A xal exi ev iS xou E Tipoc xov Z Xoyw. Xeyw 8Vj, oxi xal 
eXdyiaxoi. ei yap eiaiv oi 9, H, K, A ec;rjc dvdXoyov 
eXdyiaxoi ev xe xolc xou A Tipoc xov B xal xou T Tipoc xov 
A xal ev x£> xou E Tipoc xov Z Xoyoic, eaxoaav oi N, S, 
M, O. xal ckci eaxiv 6?6A Tipoc xov B, ouxwc 6 N Tipoc 
xov S, oi 8e A, B eXd/iaxoi, oi 8e eXdyiaxoi ^expouai xouc 
xov auxov Xoyov eyovxac iadxic o xe ^ie[£(x>v xov ^.ei^ova 
xal 6 eXdaaov xov eXdaaova, xouxeaxiv o xe fjyoujievoc 
xov f)You[ievov xal 6 CTio^evoc xov euo^evov, 6 B apa xov 
5 ^.expel. 8ia xa auxa 8r) xal 6 T xov 5 fjiexpeT - oi B, T 
apa xov S ^.expouaiv xal 6 eXdyiaxoc apa utio twv B, T 
^expou^evoc xov H [isipr\asi. eXdyiaxoc Se utio xQv B, T 
^.expeTxai 6H'6H apa xov H ^xexpeT 6 ^ei^wv xov eXdaaova - 
OTiep eaxiv d8uvxaxov. oux apa eaovxai xivec xov 9, H, K, 
A eXdaaovec dpid^oi ec;rjc ev xe xw xou A Tipoc xov B xal 
tw xou T Tipoc xov A xal exi tu xou E Tipoc xov Z XoyG. 



required to show. 

Proposition 4 

For any multitude whatsoever of given ratios, (ex- 
pressed) in the least numbers, to find the least numbers 
continuously proportional in these given ratios. 
A i ' B ' ' 

C' 1 D 1 

E ' ' F ' 1 

Ni ' H' ' 

O' ' G 

Mi 1 Ki 1 

P i 1 L i 1 

Let the given ratios, (expressed) in the least numbers, 
be the (ratios) of A to B, and of C to D, and, further, 
of E to F. So it is required to find the least numbers 
continuously proportional in the ratio of A to B, and of 
C to B, and, further, of E to F. 

For let the least number, G, measured by (both) B and 
C have be taken [Prop. 7.34]. And as many times as B 
measures G, so many times let A also measure H . And as 
many times as C measures G, so many times let D also 
measure K. And E either measures, or does not measure, 
K. Let it, first of all, measure (K). And as many times as 
E measures K, so many times let F also measure L. And 
since A measures H the same number of times that B also 
(measures) G, thus as A is to B, so H (is) to G [Def. 7.20, 
Prop. 7.13]. And so, for the same (reasons), as C (is) to 
D, so G (is) to K, and, further, as E (is) to F, so K (is) 
to L. Thus, H, G, K, L are continuously proportional in 
the ratio of A to B, and of C to D, and, further, of E to 
F. So I say that (they are) also the least (numbers con- 
tinuously proportional in these ratios) . For if H , G, K, 
L are not the least numbers continuously proportional in 
the ratios of A to B, and of C to D, and of E to F, let N, 
O, M, P be (the least such numbers) . And since as A is 
to B, so N (is) to O, and A and B are the least (numbers 
which have the same ratio as them), and the least (num- 
bers) measure those (numbers) having the same ratio (as 
them) an equal number of times, the greater (measur- 
ing) the greater, and the lesser the lesser — that is to say, 
the leading (measuring) the leading, and the following 
the following [Prop. 7.20], B thus measures O. So, for 
the same (reasons), C also measures O. Thus, B and C 
(both) measure O. Thus, the least number measured by 
(both) B and C will also measure O [Prop. 7.35]. And 
G (is) the least number measured by (both) B and C. 



231 



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ELEMENTS BOOK 8 



A' ' B ' 

T ' ' A ' 

E' ' Z ' 

N' ' ®| ' 

Hi ' H' ' 

Mi - ' K - ' 

() ' 

n> 1 

p i 1 

s | 1 

T 1 

Mrj ^sxpsixw 8r) 6 E xov K, xal siXrjcp'dw utio xcov E, 
K sA&xicttoc; \±£Tpou\±svoz dpid^ioc; 6 M. xal oadxic; \iev 
6 K xov M ^sxpsT, xoaauxdxic; xal sxdxspoc; xwv 0, H 
sxdxspov xwv N, S ^.sxpeixw, oadaxic; 8s 6 E xov M \±e- 
xpeT, xoaauxdxic; xal 6 Z xov O ^.expeixw. etc! iadxic; 6 9 
xov N ^.sxpsT xal 6 H xov S, saxiv apa 6? 6 9 Tipoc; xov 
H, ouxwc; 6 N Tipoc; xov S. 6c; 8s 6 Tipoc; xov H, ouxcoc 
6 A Tipoc; xov B - xal £><; apa 6 A Tipoc; xov B, ouxwc; 6 N 
Tipoc; xov S. 8id xa auxa hf] xal 6c; 6 T Tipoc; xov A, ouxoc; 
6 S Tipoc; xov M. udXiv, etce! iadxic; 6 E xov M ^sxpsl xal 
6 Z xov O, saxiv apa wcoE Tipoc; xov Z, ouxoc; 6 M Tipoc; 
xov 0- oi N, S, M, O apa e^r\c, dvdXoyov siaiv sv xolc; xou 
xe A Tipoc; xov B xal xou T Tipoc; xov A xal sxi xou E Ttpoc 
xov Z Xoyou;. Xsyto 8rj, oxi xal sXd)(iaxoi ev xou; A B, T 
A, E Z Xoyou;. ei yap y.r\, saovxai xivsc; xGv N, S, M, 
sXdaaovsc; dpi'dpioi Ec^fjc; dvdXoyov ev xou; A B, T A, E Z 
Xoyou;. Eaxwaav oi n, P, S, T. xal stisi saxiv «<; 6 H Tipoc; 
xov P, ouxcx; 6 A Tipoc; xov B, oi 8s A, B sXd)(iaxoi, oi 8s 
sXd)(iaxoi jisxpouai xouc; xov auxov Xoyov S)(ovxac; auxolc; 
iadxic; o xe r]you^.svo<; xov f]you^isvov xal 6 stiojisvoc; xov 
£Ti6[i£vov, 6 B apa xov P ^sxpsT. 8ia xa auxa 8r] xal 6 T 
xov P ^ExpsT- oi B, T apa xov P ^tsxpouaiv. xal 6 sXdxiaxoc; 
apa utio x<3v B, T jisxou^svoc; xov P ^sxpr|asi. sXdxiaxoc; 
8s utio x<3v B, r (jisxpoujjiEvoc; saxiv 6 H- 6 H apa xov P 
^ExpsT. xai saxiv cbc; 6 H Tipoc; xov P, ouxwc; 6 K Tipoc; xov 
£• xal 6 K apa xov S ^sxpsl. [isxpsl 8s xal 6 E xov £• oi E, 
K apa xov E [isxpouaiv. xal 6 sXdxiaxoc; apa utio xwv E, K 
^sxpou^tsvoc; xov S jiExpr]OEi. sXdxiaxoc; 8s utio xGv E, K 
[isxpou^svoc; saxiv 6 M- 6 M apa xov E [isxpsT 6 ^isiCwv xov 
sXdaaova- oTisp saxiv d8uvaxov. oux apa saovxai xivsc; xwv 



Thus, G measures O, the greater (measuring) the lesser. 
The very thing is impossible. Thus, there cannot be any 
numbers less than H, G, K, L (which are) continuously 
(proportional) in the ratio of A to B, and of C to D, and, 



further, of E to F. 

A' 1 B 

Ci 1 D 

Ei ' F 

Ni ' H 

Oi G 

Mi ' K 

pi 1 



Q 

R 

S ' ' 

T ' 

So let E not measure K. And let the least num- 
ber, M, measured by (both) E and K have been taken 
[Prop. 7.34] . And as many times as K measures M, so 
many times let H, G also measure N, O, respectively. 
And as many times as E measures M, so many times let 
F also measure P. Since H measures N the same num- 
ber of times as G (measures) O, thus as H is to G, so 
N (is) to O [Dei. 7.20, Prop. 7.13]. And as H (is) to G, 
so A (is) to B. And thus as A (is) to B, so N (is) to 
O. And so, for the same (reasons), as C (is) to D, so O 
(is) to M. Again, since E measures M the same num- 
ber of times as F (measures) P, thus as E is to F, so 
M (is) to P [Def. 7.20, Prop. 7.13]. Thus, N, O, M, P 
are continuously proportional in the ratios of A to B, and 
of C to D, and, further, of E to F. So I say that (they 
are) also the least (numbers) in the ratios of A B, C D, 
E F. For if not, then there will be some numbers less 
than N, O, M, P (which are) continuously proportional 
in the ratios of A B, C D, E F. Let them be Q, R, S, 
T. And since as Q is to R, so A (is) to B, and A and B 
(are) the least (numbers having the same ratio as them), 
and the least (numbers) measure those (numbers) hav- 
ing the same ratio as them an equal number of times, 
the leading (measuring) the leading, and the following 
the following [Prop. 7.20], B thus measures R. So, for 
the same (reasons), C also measures R. Thus, B and C 
(both) measure R. Thus, the least (number) measured by 
(both) B and C will also measure R [Prop. 7.35]. And G 
is the least number measured by (both) B and C. Thus, 
G measures R. And as G is to R, so K (is) to S. Thus, 



232 



STOIXEIQN rf. 



ELEMENTS BOOK 8 



N, S, M, O sXdaaovs<; dpi/d^iol s^fjc dvdXoyov sv xs xoTc 
toO A 7ip6<; xov B xal xoO T Ttpoc; xov A xal exi xoO E 7tp6<; 
xov Z Xoyoic oi N, S, M, O dpa e^fjc; dvdXoyov eXd)(iaxo( 
siaiv sv xou; A B, T A, E Z Xoyoic oitsp s8si SsT^ai. 



£'. 

Oi STUTts8oi dpiif)[!ot 7tp6<; dXXr|Xou<; Xoyov s/ouai xov 
auyxsi^isvov sx xwv TtXsupwv. 

A' 1 

B ' ' 

r i 1 a i 1 

E' ' Z ' 

Hi ' 

©| 1 

Kj 1 

A' ' 

"Earaaav stutcs8oi apid^ol oi A, B, xal xou [ikv A 
TtXsupal saxcoaav oi T, A dpi , d[io[, xoO 8s B oi E, Z - Xsya>, 
oxi 6 A 7ip6<; xov B Xoyov sxsi T ° v cruyxsi^isvov ex xwv 
TiXeupwv. 

Aoywv yap BoiDevxwv xoO xe ov eyei 6 T 7tp6<; xov E xal 
6 A Ttpoc xov Z eiXfjcp-dwaav dpid^oi s^<; sXd)(iaxoi sv xou; 
r E, A Z Xoyoic, oi H, 6, K, &axs slvai &><; ^isv xov T npbc, 
xov E, ouxw<; xov H Ttp6<; xov 0, <i><; Ss xov A 7tp6<; xov Z, 
ouxcoc; xov Ttpoc; xov K. xal 6 A xov E TtoXXaTcXaaidaac; 

XOV A TIOISIXCO. 

Kal lnei 6 A xov jisv T TroXXajcXaaidaac; xov A 
7ien;oir]X£v, xov Be E TioXXaTcXaaidaac; xov A iiETco[r)XSv, 
saxiv dpa 6 T npoc, xov E, oux«<; 6 A upoc; xov A. ci><; 
8s 6 r icpoc; xov E, ouxwc; 6 H icpoc; xov ©• xal cb<; dpa 6 
H Ttpoc; xov 0, ouxw<; 6 A Ttpoc; xov A. TtdXiv, STtsl 6 E xov 
A TtoXXaTtXaaidaac; xov A TiETioirjXEv, dXXa ^trjv xal xov Z 
TtoXXaTtXaaidaac; xov B TtSTtofyxsv, saxiv dpa u; 6 A Ttpoc; 
xov Z, ouxw<; 6 A Ttpoc; xov B. dXX' cbc; 6 A Ttpoc; xov Z, 
ouxwc; 6 Ttpoc; xov K- xal cbc; dpa 6 Ttpoc; xov K, ouxox 
6 A Ttpoc; xov B. eSsi^i}/] Se xal <i>c; 6 H Ttpoc; xov 0, ouxox 
6 A Ttpoc; xov A- 8i° laou dpa saxiv cbc; 6 H Ttpoc; xov K, 
[ouxck] 6 A Ttpoc; xov B. 6 8s H Ttpoc; xov K Xoyov sxsi 



if also measures S [Def. 7.20]. And E also measures 
S 1 [Prop. 7.20]. Thus, E and K (both) measure 5. Thus, 
the least (number) measured by (both) E and K will also 
measure S [Prop. 7.35]. And M is the least (number) 
measured by (both) E and K. Thus, M measures S, the 
greater (measuring) the lesser. The very thing is impos- 
sible. Thus there cannot be any numbers less than N, O, 
M, P (which are) continuously proportional in the ratios 
of A to B, and of C to D, and, further, of E to F. Thus, 
N, O, M, P are the least (numbers) continuously propor- 
tional in the ratios of A B, C D, E F. (Which is) the very 
thing it was required to show. 

Proposition 5 

Plane numbers have to one another the ratio compoun- 
ded* out of (the ratios of) their sides. 

A ' 

B 

C 1 Di 1 

E i 1 F i 1 

G ' 

Hi ' 

K 1 

L ' ' 

Let A and B be plane numbers, and let the numbers 
C, D be the sides of A, and (the numbers) E, F (the 
sides) of B. I say that A has to B the ratio compounded 
out of (the ratios of) their sides. 

For given the ratios which C has to E, and D (has) to 
F, let the least numbers, G, H, K, continuously propor- 
tional in the ratios C E, D F have been taken [Prop. 8.4], 
so that as C is to E, so G (is) to H, and as D (is) to F, so 
H (is) to K. And let D make L (by) multiplying E. 

And since D has made A (by) multiplying C, and has 
made L (by) multiplying E, thus as C is to E, so A (is) to 
L [Prop. 7.17]. And as C (is) to E, so G (is) to H. And 
thus as G (is) to H, so A (is) to L. Again, since E has 
made L (by) multiplying D [Prop. 7.16], but, in fact, has 
also made B (by) multiplying F, thus as D is to F, so L 
(is) to B [Prop. 7.17]. But, as D (is) to F, so H (is) to 
K. And thus as H (is) to K, so L (is) to B. And it was 
also shown that as G (is) to H, so A (is) to L. Thus, via 
equality, as G is to K, [so] A (is) to B [Prop. 7.14]. And 
G has to K the ratio compounded out of (the ratios of) 
the sides (of A and B). Thus, A also has to B the ratio 
compounded out of (the ratios of) the sides (of A and B). 



233 



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ELEMENTS BOOK 8 



xov auyxei^ievov ex xSv TtXeupfiv xod 6 A apex Ttpoc; xov 
B Xoyov styei xov auyxeijievov ex xfiiv TtXeupfiv ojtep e8ei 
8eT<;ai. 

t i.e., multiplied. 

f '. 

'Edv Saw ottoooiouv dtpi'djiol e<;rj<; dvdXoyov, 6 8e 
Tipoxoc; xov 8euxepov [ir] nsxpfj, oi)8e aXkoq ouSei? oi)8eva 
Hexprpei. 

A' ' 

B ' ' 

r ' ' 

A 1 ' 

Ei 1 

Z' ' 

H' 1 

©i 1 

"Eaxwaav oitoaoioOv dpid^ioi e^rjc; dvdXoyov oi A, B, 
r, A, E, 6 8e A xov B \xr] jjiexpsixw Xeyw, oxi ouSe aXXoc 
ou8ek; ouSeva [lexprjaei. 

"Oxi (iev ouv oi A, B, r, A, E zE,f\z dXXiqXouc; ou [iz- 
xpouoiv, cpavepov ou8e yap 6 A xov B ^.expeT. Xeyw 
8rj, oxi ou8e aXXoc; ou8el<; ou8eva [icxprpei. z\ yap 8u- 
vaxov, jiexpeixw 6 A xov T. xod oaoi elolv oi A, B, T, 
xoaouxoi eiXf](p'dwaav IXd/iaxoi dpid^ioi xfiv xov auxov 
Xoyov e/ovxtov xolc; A, B, T oi Z, H, 0. xod knel oi Z, 
H, O ev tu auxw X6ya> eiai xou; A, B, T, xai eaxiv laov xo 
TtXrydoc; x«v A, B, T xw TCXr^ei iwv Z, H, O, 8i' laou dpa 
eaxiv cbc 6 A 7tp6<; xov T, ouxcoc; 6 Z 7tp6<; xov 8. xod eirei 
eaxiv cbc; 6 A 7tp6<; xov B, ouxgk 6 Z Ttpoc xov H, oO ^texpel 
8e 6 A xov B, ou ^expeT dpa ouSe 6 Z xov H- oux dpa ^tovd<; 
eaxiv 6 Z- f] yap ^ovac icdvxa dpid^tov ^texpeT. xai eiaiv oi 
Z, 9 icpwxoi Tipoc; dXXr|Xou<; [ouSe 6 Z dpa xov 9 fiexpel]. 
xat eaxiv cb<; 6 Z Tipoc; xov 9, oux«<; 6 A jepoe; xov E ou8e 
6 A dpa xov T ^texpeT. o^toiog Br] Beic;o^ev, oxi ouSe aXXoc 
ou8ei<; ouSeva ^texpr]aei- oicep eSei SeT^ai. 



'Edv Saiv ouoaoioOv dpiif)[ioi [e?fj<;] dvdXoyov, 6 Be 
Tipwxog xov ea/axov (jLexpfj, xai xov Seuxepov ^expr]aei. 



(Which is) the very thing it was required to show. 



Proposition 6 

If there are any multitude whatsoever of continuously 
proportional numbers, and the first does not measure the 
second, then no other (number) will measure any other 
(number) either. 

A' 1 

Bi 1 

Ci 1 

Di 1 

Ei 1 

F i 1 

G 

Hi 1 

Let A, B, C, D, E be any multitude whatsoever of 
continuously proportional numbers, and let A not mea- 
sure B. I say that no other (number) will measure any 
other (number) either. 

Now, (it is) clear that A, B, C, D, E do not succes- 
sively measure one another. For A does not even mea- 
sure B. So I say that no other (number) will measure 
any other (number) either. For, if possible, let A measure 
C. And as many (numbers) as are A, B, C, let so many 
of the least numbers, F, G, H, have been taken of those 
(numbers) having the same ratio as A, B, C [Prop. 7.33]. 
And since F, G, H are in the same ratio as A, B, C, and 
the multitude of A, B, C is equal to the multitude of F, 
G, H, thus, via equality, as A is to C, so F (is) to H 
[Prop. 7.14]. And since as A is to B, so F (is) to G, 
and A does not measure B, F does not measure G either 
[Def. 7.20]. Thus, F is not a unit. For a unit measures 
all numbers. And F and H are prime to one another 
[Prop. 8.3] [and thus F does not measure H]. And as 
F is to H, so A (is) to C. And thus A does not measure 
C either [Def. 7.20]. So, similarly, we can show that no 
other (number) can measure any other (number) either. 
(Which is) the very thing it was required to show. 

Proposition 7 

If there are any multitude whatsoever of [continu- 
ously] proportional numbers, and the first measures the 



234 



STOIXEIQN rf. 



ELEMENTS BOOK 8 



A' ' 

B 1 

V 1 

A> 1 

'Eoxcooav onoaoiouv dpid^iol e^fjc; dvdXoyov ol A, B, T, 
A, 6 Se A xov A [isxperlxw Xeyto, oxi xal 6 A xov B jxexpel. 

El yap ou (iexpsT 6 A xov B, ou8e dXXoc; oOSeu; ou8eva 
\L£Tpx]aev jjiexpeT 8e 6 A xov A. fjisxpeT dpa xal 6 A xov B- 
oTterp e8ei 8eT?ai. 



*]'• 

'Edv 8uo dpii9|ji«v jiexa^u xaxd xo auve)(ec; dvdXoyov 
e^miKxoaiv dpi%o£, oaoi sic; auxouc; |iexa^u xaxd xo au- 
v£)(£<; dvoXoyov e^ranxouaiv dpidjiol, xooouxoi xal ek xouc 
xov auxov Xoyov e^ovxac; [auxolc;] ^£xa<;u xaxd xo ouve)(ec; 
dvdXoyov 6[i7tEaouvxai 

A i ' E' ' 

r ' ' M' ' 

A ' N' ' 

B ' ' Z' ' 

H ' 

O' ' 

K 1 ' 

A' ' 

Auo yap dpn5^Gv xwv A, B jisxac^u xaxd xo auvexec; 
dvdXoyov e^mmxexwaav dpid|iol ol T, A, xal 7i£Tioir]o , da) d>c 
6 A Ttpoc; xov B, ouxcoc; 6 E npoc; xov Z- Xsyw, oxi oaoi sic, 
xouc; A, B ^iexac;u xaxd xo auve)(£<; dvdXoyov e^msTtxtoxaaiv 
dpid^oi, xoaouxoi xal zlc. xoug E, Z ^exac^u xaxd xo auvzyzz 
dvdXoyov e^Tteaouvxai. 

"Oaoi ydp rial xcp 7tXrji!tei ol A, B, T, A, xooouxoi 
riXr|cpiL>Gjaav sXd)(iaxoi dpiiSjiol xfiv xov auxov Xoyov 
s/ovxiov xoTc; A, r, A, B ol H, 9, K, A- ol dpa axpoi 
auxwv ol H, A npGxoi upoc dXXf|Xouc; elalv. xal enel ol A, 
r, A, B xolc; H, 0, K, A ev xa> auxco Xoyio elalv, xal eaxiv 
I'oov xo TiXfj'doc; xGv A, T, A, B xG txXt^t&si xGv H, 9, K, 
A, 8i° I'oou dpa eaxlv u;6 A Ttpoc; xov B, ouxgjc; 6 H npoc; 
xov A. &>c, 8e 6 A npoc; xov B, ouxtog 6 E npoc; xov Z- xal 



last, then (the first) will also measure the second. 

A' ' 

B' 1 

C' 1 

Di 1 

Let A, B, C, D be any number whatsoever of continu- 
ously proportional numbers. And let A measure D. I say 
that A also measures B. 

For if A does not measure B then no other (number) 
will measure any other (number) either [Prop. 8.6]. But 
A measures D. Thus, A also measures B. (Which is) the 
very thing it was required to show. 

Proposition 8 

If between two numbers there fall (some) numbers in 
continued proportion then, as many numbers as fall in 
between them in continued proportion, so many (num- 
bers) will also fall in between (any two numbers) having 
the same ratio [as them] in continued proportion. 

A i 1 E i 1 

Ci 1 M' 1 

D' ' N' ' 

Bi 1 F ' ' 

G' ' 

H' ' 

K' ' 

L ' ' 

For let the numbers, C and D, fall between two num- 
bers, A and B, in continued proportion, and let it have 
been contrived (that) as A (is) to B, so E (is) to F. I say 
that, as many numbers as have fallen in between A and 
B in continued proportion, so many (numbers) will also 
fall in between E and F in continued proportion. 

For as many as A, B, C, D are in multitude, let so 
many of the least numbers, G, H, K, L, having the same 
ratio as A, B, C, D, have been taken [Prop. 7.33]. Thus, 
the outermost of them, G and L, are prime to one another 
[Prop. 8.3]. And since A, B, C, D are in the same ratio 
as G, H, K, L, and the multitude of A, B, C, D is equal 
to the multitude of G, H, K, L, thus, via equality, as A is 
to B, so G (is) to L [Prop. 7.14]. And as A (is) to B, so 



235 



STOIXEIQN rf. 



ELEMENTS BOOK 8 



tbc; dpa 6 H Ttpoc; tov A, outmc; 6 E Ttpoc; tov Z. ol 6e H, A 
TtpfiToi, ol 8e upwxoi xal eXdxioToi, ol 8e eXdxioToi dpi'diiol 
(iETpouai touc; tov auTov Xoyov exovTac iadxic; 6 tc iieiCiov 
xov [iei^ova xal 6 eXdaatov tov eXdaaova, TouTeaTiv 6 Te 
fjyouijievoc; tov fjyouiievov xal 6 eTtojievoc; tov eTiojievov. 
iadxic; apa 6 H tov E ^iCTpeT xal 6 A tov Z. oadxic 8f] 6 H 
tov E [Lsxpei, TooauTaxic; xal exaTcpoc; to>v 0, K exaTepov 
tGv M, N LiETpehto- oi H, 9, K, A apa touc; E, M, N, Z 
iadxic; [iETpouaiv. oi H, 6, K, A apa tou; E, M, N, Z ev tG 
auTfi Xoycp eiaiV dXXd ol H, O, K, A toTc; A, T, A, B ev t£> 
auTW Xoyio eiaiv xal ol A, T, A, B apa toTc; E, M, N, Z ev 
tw auTcp Xoycp eiaiv. oi 8e A, T, A, B eJ;rjc; dvdXoyov eiaiv 
xal oi E, M, N, Z apa e^fjg dvdXoyov eiaiv. oaoi apa eic; 
touc; A, B ^iCTac;u xaTa to auvexec; dvdXoyov eiineiiTtoxaaiv 
dpnSiioi, ToaouToi xal eic; touc; E, Z [ieTa^u xaTa to auvexec; 
dvdXoyov eii7ien;T«xaaiv dpid^oi- orcep e8ei 8eTc;ai. 



E (is) to F. And thus as G (is) to L, so E (is) to F. And 
G and L (are) prime (to one another). And (numbers) 
prime (to one another are) also the least (numbers hav- 
ing the same ratio as them) [Prop. 7.21]. And the least 
numbers measure those (numbers) having the same ratio 
(as them) an equal number of times, the greater (measur- 
ing) the greater, and the lesser the lesser — that is to say, 
the leading (measuring) the leading, and the following 
the following [Prop. 7.20]. Thus, G measures E the same 
number of times as L (measures) F. So as many times as 
G measures E, so many times let H, K also measure M, 
N, respectively. Thus, G, H, K, L measure E, M, N, 
F (respectively) an equal number of times. Thus, G, H, 
K, L are in the same ratio as E, M, N, F [Def. 7.20] . 
But, G, H, K, L are in the same ratio as A, C, D, B. 
Thus, A, C, D, B are also in the same ratio as E, M, N, 
F. And A, C, D, B are continuously proportional. Thus, 
E, M, N, F are also continuously proportional. Thus, 
as many numbers as have fallen in between A and B in 
continued proportion, so many numbers have also fallen 
in between E and F in continued proportion. (Which is) 
the very thing it was required to show. 



Proposition 9 



'Edv 8uo dpii9|jiol TtpwToi npoc; dXXrjXouc; Saiv, xal 
eic; auTouc; ^CTac;u xaTa to auvexec; dvdXoyov e^TUKToaiv 
dprd^toi, oaoi eic; auTouc; ^CTa^u xaTa to auvexec; dvdXoyov 
eiiTUTiTouaiv dpid^toi, ToaoOroi xal exaTepou auT«v xal 
iiovd8o<; iieTac^u xaTa to auve/ec; dvdXoyov e^uieaouvTai. 

A i 1 ©i 1 



r i- 



K> 
Ah 



B h 

Eh 



O 



'EaTwaav 8uo dpid^ol KpWTOi Ttpoc; dXXrjXouc; oi A, 
B, xal eic; auTouc; ^.eTa^u xaTa to auvexec; dvdXoyov 
e^TUKTCToaav oi T, A, xal exxeiadw f) E ^tovdc;- Xeyw, 
oti oaoi eic touc; A, B ^CTac^u xaTa to auvexec; dvdXoyov 
e^meTiTWxaaiv dpid^ioi, ToaouToi xal exaTepou t«v A, 
B xal xfjc; jiovdBoc; iieTa^u xaTa to auvexec; dvdXoyov 
eiiTieaouvTai. 

EiX^cp-dwaav yap 8uo (iev dpi'diiol eXdxioToi ev t£3 twv 
A, T, A, B Xoycp ovtcc; oi Z, H, TpeTc; 8e oi <d, K, A, xal del 



If two numbers are prime to one another and there 
fall in between them (some) numbers in continued pro- 
portion then, as many numbers as fall in between them 
in continued proportion, so many (numbers) will also fall 
between each of them and a unit in continued proportion. 

A i 1 H 1 



K 



Dh 



F i 1 



O 



G 1 P 1 

Let A and B be two numbers (which are) prime to 
one another, and let the (numbers) C and D fall in be- 
tween them in continued proportion. And let the unit E 
be set out. I say that, as many numbers as have fallen 
in between A and B in continued proportion, so many 
(numbers) will also fall between each of A and B and 
the unit in continued proportion. 

For let the least two numbers, F and G, which are in 
the ratio of A, C, D, B, have been taken [Prop. 7.33]. 



236 



STOIXEIQN rf. 



ELEMENTS BOOK 8 



sSfjc; evl tiXeiouc;, ecoc; dv taov yevrjxai to TCXrydoc; auxfiiv xfi 
TtXn^T&Ei xfiv A, T, A, B. siXr]cpi9«oav, xal eaxcoaav ol M, N, 
S, O. cpavspov 8r), oxi 6 [isv Z sauxov 7toXXaTtXaaidaa<; xov 
TCSTioirjXEv, xov Ss TroXXaTtXaaidaac; xov M tiettoi/jxev, 
xal 6 H eauxov (isv rcoXXaTiXaaidaac; xov A TtETtoirjxsv, xov 
8e A TtoXXanXaaidaac; xov O TtETtoirjxsv. xal etc! oi M, N, 
S, O eXd)(iaxo[ siai xtov xov auxov Xoyov e/ovxwv xoT<; Z, 
H, eial 6s xal oi A, T, A, B sXa/ioxoi xfiv xov auxov Xoyov 
exovxcov xou; Z, H, xai saxiv laov xo TxXfjiE>oc; xwv M, N, S, 
O xG TtXr] n &eL xwv A, V, A, B, sxaaxoc; dpa xwv M, N, S, O 
exdaxw xwv A, T, A, B lao? eaxiv l'ao<; dpa saxlv 6 [isv M 
iw A, 6 8e O iw B. xal etie! 6 Z eauxov TroXXaTtXaaidaac; 
xov O KETXoirjxev, 6 Z dpa xov ^lexpei xaxa xa<; ev xw Z 
^iovd8a<;. [icxpel 8e xal f\ E ^tova<; xov Z xaxa xd<; sv auxw 
^ovd8a<;- ladxic; dpa f] E ^tovdc; xov Z apidjjiov jiexpsT xal 6 Z 
xov O. eaxiv dpa &>q rj E (iova<; 7tp6<; xov Z dpid^tov, ouxoc 
6 Z Kp6<; xov O. ndXiv, end 6 Z xov 9 TtoXXaTtXaaidaac; 
xov M TienoirjXEv, 6 dpa xov M piexpeT xaxa xdc sv xw Z 
^iovd8a<;. jiexpeT 8e xal f] E (lovag xov Z dpid^iov xaxa xa<; ev 
auxo ^iovd8a<;- ladxic; dpa f) E fiovdc; xov Z dpid^iov ^xexpeT 
xal 6 xov M. eaxiv dpa w<; f\ E |iova<; 7ip6<; xov Z dpid^iov, 
ouxcoc; 6 O 7ip6<; xov M. £8ei)fdr] 8e xal cb<; f) E fiovdc; rcpoc; 
xov Z dpid^iov, ouxw<; 6 Z 7ip6<; xov 8' xal «<; dpa f] E fiovdc; 
Ttpoc; xov Z dpi-djiov, ouxw<; 6 Z Ttpoc; xov xal 6 icpoc; 
xov M. iao<; 8e 6 M x£) A - eaxiv dpa (be; f] E fiovdc; Ttpoc; xov 
Z dpidfiov, ouxwc; 6 Z Ttpoc; xov xal 6 Ttpoc; xov A. 8id 
xd auxd 8r] xal 6<; f] E [iovac; Ttpoc; xov H dpid^iov, ouxgx 6 
H Ttpoc; xov A xal 6 A Ttpoc; xov B. oaoi dpa ei? xouc; A, B 
fiexac;u xaxa xo auve^ec; dvdXoyov efjuteTtxcoxaaiv dpi-dfioi, 
xoaouxoi xal exaxepou xwv A, B xal (iovdSoc; xrjc; E fiexac;u 
xaxa xo auvexec; dvdXoyov e(iTteTtxoxaaiv dpi-dfior OTtep e8ei 
8eTc;ai. 



And the (least) three (numbers), H, K, L. And so on, 
successively increasing by one, until the multitude of the 
(least numbers taken) is made equal to the multitude of 
A, C, D, B [Prop. 8.2]. Let them have been taken, and 
let them be M, N, O, P. So (it is) clear that F has made 
H (by) multiplying itself, and has made M (by) multi- 
plying H . And G has made L (by) multiplying itself, and 
has made P (by) multiplying L [Prop. 8.2 corr.]. And 
since M, N, O, P are the least of those (numbers) hav- 
ing the same ratio as F, G, and A, C, D, B are also the 
least of those (numbers) having the same ratio as F, G 
[Prop. 8.2], and the multitude of M, N, O, P is equal 
to the multitude of A, C, D, B, thus M, N, O, P are 
equal to A, C, D, B, respectively. Thus, M is equal to 
A, and P to B. And since F has made H (by) multiply- 
ing itself, F thus measures H according to the units in F 
[Def. 7.15]. And the unit E also measures F according to 
the units in it. Thus, the unit E measures the number F 
as many times as F (measures) H. Thus, as the unit E is 
to the number F, so F (is) to H [Def. 7.20]. Again, since 
F has made M (by) multiplying H, H thus measures M 
according to the units in F [Def. 7.15]. And the unit E 
also measures the number F according to the units in it. 
Thus, the unit E measures the number F as many times 
as H (measures) M. Thus, as the unit E is to the number 
F, so H (is) to M [Prop. 7.20]. And it was shown that as 
the unit E (is) to the number F, so F (is) to H. And thus 
as the unit E (is) to the number F, so F (is) to H, and H 
(is) to M. And M (is) equal to A. Thus, as the unit E is 
to the number F, so F (is) to H, and H to A. And so, for 
the same (reasons), as the unit E (is) to the number G, 
so G (is) to L, and L to B. Thus, as many (numbers) as 
have fallen in between A and B in continued proportion, 
so many numbers have also fallen between each of A and 
B and the unit E in continued proportion. (Which is) the 
very thing it was required to show. 



i . 

'Edv Suo dpid^Gv exaxepou xal ^iovd8o<; jiexa^u xaxa 
xo auvEXE? dvdXoyov eututixcoctiv dpid^toi, oaoi sxaxspou 
auxwv xal jiovd8o^ \Lzxa£,b xaxa xo auvE)(E<; dvdXoyov 
s^TUTtxouaiv dpid^toi, xoaouxoi xal sic; auxouc [isxa$u xaxa 
xo auvE^EC dvdXoyov E^msaouvxai. 

Auo yap dpi^wv xwv A, B xal [iovdSoc xfjc T \ie- 
xa^u xaxa xo auve)(s<; dvdXoyov EjjiTUTixExcoaav &pid|jioi ol 
xe A, E xal oi Z, H- Xsyw, oxi oaoi exaxspou xwv A, 
B xal (jiovd8o<; xfjc T [isxa^u xaxa xo auvs)(£<; dvdXoyov 
E^KETixcoxaaiv dpvd\Lol, xoaouxoi xal zlc, xou<; A, B ^.sxa^u 
xaxa xo auvE^EC dvdXoyov suTisaouvxai. 



Proposition 10 

If (some) numbers fall between each of two numbers 
and a unit in continued proportion then, as many (num- 
bers) as fall between each of the (two numbers) and the 
unit in continued proportion, so many (numbers) will 
also fall in between the (two numbers) themselves in con- 
tinued proportion. 

For let the numbers D, E and F, G fall between the 
numbers A and B (respectively) and the unit C in con- 
tinued proportion. I say that, as many numbers as have 
fallen between each of A and B and the unit C in contin- 
ued proportion, so many will also fall in between A and 
B in continued proportion. 



237 



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ELEMENTS BOOK 8 



r — ' r — ' 

A' ' Z ' 

i: — i ii ' 

A 1 ' B ' ' 

©i i 

K 1 ' 

A' ' 

O A y&P TOV Z TioXXaTiXaaidaac; xov tioicixo, 
exdxepoc; 8e iwv A, Z xov 6 TioXXaTiXaaidaac; exdxepov 

XO)V K, A TCOIEITM. 

Kod ctici eaxiv (be; rj T [iovdc; Tipoc; tov A dpid^iov, ouxox 
6 A Tipoc; xov E, ladxic; dpa f] T [lovac; xov A dpid^iov ^.expel 
xal 6 A xov E. f) 8e r jiovdc; xov A dpid^tov ^.expeT xaxd xdc; 
ev tu A [iovd8ac xal 6 A dpa dpnf)[i6<; xov E ^xexpeT xaxd 
xdc; ev xw A ^.ovdSac;' 6 A dpa eauxov TioXXaTiXaaidaac; xov 
E -rcsTCofyxev. TidXiv, eTiei eaxiv (be; f) T [[lovdc;] Tipoc; xov 
A dpn)^6v, oux«c; 6 E Tipoc; xov A, ladxic; dpa f] T \±ovaz 
xov A dpi/d^iov ^.expeT xal 6 E xov A. f) 8e T [iovdc; xov 
A dpi-djiov ^.expel xaxd xdc; ev xw A ^.ovdBac;- xal 6 E dpa 
xov A fjiexpeT xaxd xdc; ev xw A [iovd5a<;- 6 A dpa xov E 
TioXXaTiXaaidaac; xov A 7i£Tio(r]X£v. Bid xd auxd 8r] xal 6 
\ie\) Z eauxov TioXXaTiXaaidaac; xov H TiCTio(r]xev, xov 6e H 
TioXXaTiXaaidaac; xov B Ttenoirjxev. xal excel 6 A eauxov [iev 
TioXXaTiXaaidaac; xov E TiCTioirjxev, xov Be Z TioXXaTiXaaidaac; 
xov TteTioirjxev, eaxiv dpa q;6 A Tipoc; xov Z, ouxoc; 6 E 
Tipoc; xov 0. 8id xd auxd Srj xal (be; 6 A Tipoc; xov Z, ouxwc; 6 
Tipoc; xov H. xal (be; dpa 6 E Tipoc; xov 0, ouxmc; 6 G Tipoc; 
xov H. TidXiv, CTiel 6 A exdxepov xwv E, TioXXaTiXaaidaac; 
exdxepov xwv A, K TiCTioirjxev, eaxiv dpa w;6E Tipoc; xov 0, 
ouxcoc; 6 A Tipoc; xov K. dXX' (be; 6 E Tipoc; xov 0, ouxoc; 6 A 
Tipoc; xov Z - xal (i><; dpa 6 A Tipoc; xov Z, ouxwc; 6 A Tipoc; xov 
K. TidXiv, ensl exdxepoc xwv A, Z xov TioXXaTiXaaidaac; 
exdxepov xwv K, A TiCTioirjxev, eaxiv dpa (be; 6 A Tipoc; xov 
Z, ouxok 6 K Tipoc; xov A. dXX' (be; 6 A Tipoc; xov Z, ouxwc; 6 
A Tipoc; xov K- xal (be; dpa 6 A Tipoc; xov K, ouxwc; 6 K Tipoc; 
xov A. exi eTiel 6 Z exdxepov xwv 0, H TioXXaTiXaaidaac; 
exdxepov xwv A, B TiCTioirjxev, eaxiv dpa (be; 6 Tipoc; xov 
H, ouxwc; 6 A Tipoc; xov B. (be; 5e 6 Tipoc; xov H, ouxwc; 
6 A Tipoc; xov Z' xal (be; dpa 6 A Tipoc; xov Z, ouxwc; 6 A 
Tipoc; xov B. e8e[)fdr] 8e xal (be; 6 A Tipoc; xov Z, ouxwc; o xe 
A Tipoc; xov K xal 6 K Tipoc; xov A' xal (be; dpa 6 A Tipoc; 
xov K, ouxwc; 6 K Tipoc; xov A xal 6 A Tipoc; xov B. oi A, 
K, A, B dpa xaxd xo auvexec; e$fjc; eiaiv dvdXoyov. oaoi 
dpa exaxepou xwv A, B xal xrjc; T ^.ovdBoc; jiexac;u xaxd 
xo auve^ec; dvdXoyov e^TUTixouaiv dpid^toi, xoaouxoi xal eic; 
xouc; A, B ^icxa^u xaxd xo auvexec; epmeaoOvxai- oiiep e8ei 



O — ' C' — ' 

D F i 1 

E i 1 G' 1 

A i 1 B i 1 

Hi ' 

K 

L ' ' 

For let D make H (by) multiplying F. And let D, F 
make K, L, respectively, by multiplying H . 

As since as the unit C is to the number D, so D (is) to 

E, the unit C thus measures the number D as many times 
as D (measures) E [Def. 7.20]. And the unit C measures 
the number D according to the units in D. Thus, the 
number D also measures E according to the units in D. 
Thus, D has made E (by) multiplying itself. Again, since 
as the [unit] C is to the number D, so E (is) to A, the 
unit C thus measures the number D as many times as E 
(measures) A [Def. 7.20]. And the unit C measures the 
number D according to the units in D. Thus, E also mea- 
sures A according to the units in D. Thus, D has made 
A (by) multiplying E. And so, for the same (reasons), F 
has made G (by) multiplying itself, and has made B (by) 
multiplying G. And since D has made E (by) multiplying 
itself, and has made H (by) multiplying F, thus as D is to 

F, so E (is) to H [Prop 7.17]. And so, for the same rea- 
sons, as D (is) to F, so H (is) to G [Prop. 7.18]. And thus 
as E (is) to H, so H (is) to G. Again, since D has made 
A, K (by) multiplying E, H, respectively, thus as E is to 
H, so A (is) to K [Prop 7.17]. But, as E (is) to H, so D 
(is) to F. And thus as D (is) to F, so A (is) to K. Again, 
since D, F have made K, L, respectively, (by) multiply- 
ing H, thus as D is to F, so K (is) to L [Prop. 7.18]. But, 
as D (is) to F, so A (is) to K. And thus as A (is) to K, 
so K (is) to L. Further, since F has made L, B (by) mul- 
tiplying H, G, respectively, thus as H is to G, so L (is) to 
B [Prop 7.17]. And as H (is) to G, so D (is) to F. And 
thus as D (is) to i^, so L (is) to £?. And it was also shown 
that as D (is) to F, so A (is) to AT, and K to i. And thus 
as A (is) to K, so if (is) to L, and i to £?. Thus, A, K, 
L, B are successively in continued proportion. Thus, as 
many numbers as fall between each of A and B and the 
unit C in continued proportion, so many will also fall in 
between A and B in continued proportion. (Which is) 
the very thing it was required to show 



238 



STOIXEIQN rf. 



ELEMENTS BOOK 8 



8eT<;ai. 



ia'. 

Auo xexpaycbvcov dpnD^aiv zlc, [isaoc; dvdXoyov saxiv 
dprd^ioc;, xdi 6 xsxpdyiovoc; Ttpoc; xov xexpdywvov 81- 
TtXamova Xoyov *F e P ^ ^Xsupd Ttpoc; x/]v TtXeupdv. 

A i ' 

B i 1 

r i 1 a i 1 

Ei 1 

"Eaxwaav xexpdywvoi dpid^iol oi A, B, xdi xoO [lev A 
TtXeupd eaxio 6 T, xoO 8s B 6 A- Xeyio, oxi xfiv A, B eTc; 
[isaoc, dvdXoyov eaxiv dpidjjioc;, xdi 6 A Ttpoc; xov B 8i- 
TtXamova Xoyov zyz\ fjnep 6 T Ttpoc; xov A. 

O r yap xov A TtoXXaitXaaidaac; xov E Ttoieixw. xdi 
etieI xexpdywvoc; eaxiv 6 A, TtXeupd 8e auxou eaxiv 6 T, 6 T 
dpa eauxov TtoXXaTtXaaidaac; xov A TteTtoirjxev. 8id xd auxd 
6r] xdi 6 A eauxov TtoXXaTtXaaidaac; xov B TteTtoirjxev. CTtel 
ouv 6 r exdxepov xSv T, A TtoXXaTtXaaidaac; exdxepov xSv 
A, E TteTtoirjxev, eaxiv dpa 6 T Ttpoc; xov A, ouxioc; 6 A 
Ttpoc; xov E. 6ia xd auxd 8f) xal &>c, 6 T Ttpoc; xov A, ouxioc; 6 
E Ttpoc; xov B. xal tbc; dpa 6 A Ttpoc; xov E, ouxcoc; 6 E Ttpoc; 
xov B. xaiv A, B dpa eTc; jjiaoc; dvdXoyov eaxiv dpi'djioc;. 

Aeyw 8rj, oxi xal 6 A Ttpoc; xov B BiTtXaaiova Xoyov e)(ei 
f)Tiep 6 r Ttpoc xov A. ETiel yap xpelc; dpi'djiol dvdXoyov eiaiv 
ol A, E, B, 6 A dpa Ttpoc; xov B BiTtXaaiova Xoyov eyei f\Kep 
6 A Ttpoc; xov E. uc; 8e 6 A Ttpoc; xov E, ouxwc; 6 T Ttpoc; 
xov A. 6 A dpa Ttpoc; xov B BiTtXaaiova Xoyov e)(ei fjitep f) 
T TtXeupd Ttpoc; x/]v A- oTtep eSei Belial. 



Proposition 11 

There exists one number in mean proportion to two 
(given) square numbers.^ And (one) square (number) 
has to the (other) square (number) a squared^ ratio with 
respect to (that) the side (of the former has) to the side 
(of the latter). 

A' ' 

B " 1 

O 1 D' 1 

Ei 1 

Let A and B be square numbers, and let C be the side 
of A, and D (the side) of B. I say that there exists one 
number in mean proportion to A and B, and that A has 
to B a squared ratio with respect to (that) C (has) to D. 

For let C make E (by) multiplying D. And since A is 
square, and C is its side, C has thus made A (by) multi- 
plying itself. And so, for the same (reasons), D has made 
B (by) multiplying itself. Therefore, since C has made A, 
E (by) multiplying C, D, respectively, thus as C is to D, 
so A (is) to E [Prop. 7.17]. And so, for the same (rea- 
sons), as C (is) to D, so E (is) to B [Prop. 7.18]. And 
thus as A (is) to E, so E (is) to B. Thus, one number 
(namely, E) is in mean proportion to A and B. 

So I say that A also has to B a squared ratio with 
respect to (that) C (has) to D. For since A, E, B are 
three (continuously) proportional numbers, A thus has 
to B a squared ratio with respect to (that) A (has) to E 
[Def. 5.9]. And as A (is) to E, so C (is) to D. Thus, A has 
to B a squared ratio with respect to (that) side C (has) 
to (side) D. (Which is) the very thing it was required to 
show. 



t In other words, between two given square numbers there exists a number in continued proportion. 
* Literally, "double". 



IP'. 

Auo xupwv dpiduov Buo ^xeaoi dvdXoyov eiaiv dpi%o(, 
xal 6 xu(3o<; Ttpoc xov xupov xpiTtXaaiova Xoyov ^y^l Wep f\ 
TtXeupd Ttpoc; xr)v TtXeupdv. 

TEaxoaav xupoi dpi-dpiol oi A, B xal xou ^iev A TtXeupd 
eaxco 6 T, xou 8s B 6 A- Xeyw, oxi xGv A, B Buo ^eaoi 
dvdXoyov eiaiv dpi-djioi, xal 6 A Ttpoc; xov B xpiTtXaaiova 
Xoyov e/ei fjTtep 6 T Ttpoc; xov A. 



Proposition 12 

There exist two numbers in mean proportion to two 
(given) cube numbers. t And (one) cube (number) has to 
the (other) cube (number) a cubed^ ratio with respect 
to (that) the side (of the former has) to the side (of the 
latter). 

Let A and B be cube numbers, and let C be the side 
of A, and D (the side) of B. I say that there exist two 
numbers in mean proportion to A and B, and that A has 



239 



STOIXEIQN rf. 



ELEMENTS BOOK 8 



A i 1 Ei 1 

B i 1 Zi 1 

r i 1 Hi 1 

A i 1 ©i 1 

K' ' 

'O yap r eauxov [iev TioXXauXaaidaac; xov E tioicixgj, 
xov 8e A TtoXXauXaaidaac; tov Z Tioieixo, 6 8e A eauxov 
TtoXXauXaaidaac; xov H Ttoieixw, exdxepoc; 8e xwv T, A xov 
Z TtoXXaTtXaaidaac; exdxepov xwv 0, K Ttoieixo. 

Kod etxsl xu|3oc; eaxiv 6 A, TiXeupa 8e auxou 6 T, xal 6 
r srauxov ^.ev TtoXXauXaaidaac; xov E Ttcnoirjxev, 6 T apa 
eauxov (iev TtoXXaTtXaaidaac; xov E TteTtoir]xev, xov 8e E 
TtoXXaTiXaaidaac; xov A TteTiofyxev. Bid xd auxa Sr] xal 6 
A eauxov ^iev TtoXXaTtXaaidaac; xov H TteTtoir)xev, xov 8e H 
TtoXXaTtXaaidaac; xov B TteTtoirjxev. xal eitei 6 T exdxepov 
iwv T, A TtoXXaTtXaaidaac; exdxepov xwv E, Z TteTtoirjxev, 
eaxiv apa (be; 6 T Ttpoc; xov A, ouxwc; 6 E Ttpoc; xov Z. 
8id xd auxa 8f) xal (be; 6 T Ttpoc; xov A, ouxwc; 6 Z Ttpoc; 
xov H. TtdXiv, euel 6 T exdxepov xQv E, Z TtoXXaTtXaaidaac; 
exdxepov xQv A, TteTtoir]xev, eaxiv apa (be; 6 E Ttpoc; xov 
Z, oux«<; 6 A Ttpoc; xov 6. (be; 8e 6 E Ttpoc; xov Z, ouxwc; 6 T 
Ttpoc; xov A- xal (be; apa 6 T Ttpoc; xov A, ouxwc; 6 A Ttpoc; xov 
0. TtdXiv, CTtel exdxepoc; xwv T, A xov Z TtoXXaTtXaaidaac; 
exdxepov xwv 0, K TteTtoirjxev, eaxiv apa u?6T Ttpoc; xov 
A, ouxwc; 6 Ttpoc; xov K. TtdXiv, CTtel 6 A exdxepov iSv 
Z, H TtoXXaTtXaaidaac; exdxepov xwv K, B TteTtoirjxev, eaxiv 
apa (be 6 Z Ttpoc; xov H, ouxw<; 6 K Ttpoc; xov B. cLk 8e 6 Z 
Ttpoc; xov H, ouxwc; 6 T Ttpoc; xov A- xal (be; apa 6 T Ttpoc; 
xov A, ouxwc; o xe A Tipoc; xov xal 6 Ttpoc; xov K xal 
6 K Tipoc; xov B. xebv A, B apa 8uo ueaoi dvdXoyov eiaiv 
oi 0, K. 

Aeyw 8rj, oxi xal 6 A Ttpoc; xov B xpiTtXaaiova Xoyov e^ei 
r]Ttep 6 r Tipoc; xov A. euel yap xeaaapec; dpid^iol dvdXoyov 
eiaiv oi A, 0, K, B, 6 A apa Tipoc; xov B xpiTtXaaiova Xoyov 
e^ei f)Tiep 6 A Tipoc; xov 0. (be; Se 6 A Tipoc; xov 0, ouxoc; 6 
T Tipoc; xov A- xal 6 A [apa] Tipoc; xov B xpiTtXaaiova Xoyov 
exei fjnep 6 T Tipoc; xov A- oTtep e8ei 8eTc;ai. 

t In other words, between two given cube numbers there exist two nun 
t Literally, "triple". 



to B a cubed ratio with respect to (that) C (has) to D. 
A i 1 E i 1 

B i 1 F i 1 

Ci 1 Gi 1 

Di 1 Hi 1 

Ki 1 

For let C make E (by) multiplying itself, and let it 
make F (by) multiplying D. And let D make G (by) mul- 
tiplying itself, and let C, D make H, K, respectively, (by) 
multiplying F. 

And since A is cube, and C (is) its side, and C has 
made E (by) multiplying itself, C has thus made E (by) 
multiplying itself, and has made A (by) multiplying E. 
And so, for the same (reasons), D has made G (by) mul- 
tiplying itself, and has made B (by) multiplying G. And 
since C has made E, F (by) multiplying C, D, respec- 
tively, thus as C is to D, so E (is) to F [Prop. 7.17]. And 
so, for the same (reasons), as C (is) to D, so F (is) to G 
[Prop. 7.18]. Again, since C has made A, H (by) multi- 
plying E, F, respectively, thus as E is to F, so A (is) to 
H [Prop. 7.17]. And as E (is) to F, so C (is) to D. And 
thus as C (is) to D, so A (is) to H. Again, since C, D 
have made iJ, K, respectively, (by) multiplying F, thus 
as C is to D, so _ff (is) to K [Prop. 7.18]. Again, since D 
has made K, B (by) multiplying F, G, respectively, thus 
as F is to G, so K (is) to B [Prop. 7.17]. And as F (is) 
to G, so C (is) to £>. And thus as C (is) to D, so A (is) 
to H, and to i\T, and K to B. Thus, iJ and K are two 
(numbers) in mean proportion to A and B. 

So I say that A also has to B a cubed ratio with re- 
spect to (that) C (has) to D. For since A, H, K, B are 
four (continuously) proportional numbers, A thus has 
to B a cubed ratio with respect to (that) A (has) to H 
[Def. 5.10]. And as A (is) to H, so C (is) to D. And 
[thus] A has to B a cubed ratio with respect to (that) C 
(has) to D. (Which is) the very thing it was required to 
show. 



in continued proportion. 



'Edv fiaiv oaoiSrjTioxouv dpid^tol e^fjc; dvdXoyov, xal 
TioXXaTiXaaidaac; exaaxoc eauxov Tioi/j xiva, oi yevojievoi 
ec; auxwv dvdXoyov eaovxai' xal edv oi ec; dp^fjc xouc 
yevo^ievouc; TtoXXauXaaidaavxec; Ttoiwai xivac;, xal auxol 
dvdXoyov eaovxai [xal del Ttepl xouc; axpouc; xouxo au^paivei] . 

'Eaxwaav oTioaoiouv dpid^iol ec^fjc; dvdXoyov, oi A, B, 



Proposition 13 

If there are any multitude whatsoever of continuously 
proportional numbers, and each makes some (number 
by) multiplying itself, then the (numbers) created from 
them will (also) be (continuously) proportional. And if 
the original (numbers) make some (more numbers by) 
multiplying the created (numbers) then these will also 



240 



STOIXEIQN rf. 



ELEMENTS BOOK 8 



r, q; 6 A Ttpoc; xov B, ouxioc; 6 B 7tp6<; xov T, xod oi 
A, B, r eauxouc; [iev TToXXaitXaaLaaavxsc; xoug A, E, Z 
noieixtoaav, xou<; Se A, E, Z TtoXXaTtXaaidaavxec; xou<; H, 
9, K Tioisraoaav Xeyw, ° Tl of Te A, E, Z xal oi H, 9, K 
zZjf\Z dvdXoyov sioiv. 



A 
B 

r 



M 
N 
O 

n 



A 
E 
Z 

H 


K 

^iev yap A xov B TCoXXanXamdaac; xov A toieixco, 
sxdxspog 8e xwv A, B xov A 7ioXXaTtXaaidaa<; exdxepov iSv 
M, N Tioieixw. xal TtdXiv 6 \±kv B xov T noXXajiXaaidaa^ xov 
S Ttoisixio, sxdxspog 8s xfiv B, T xov S TTOXXanXaaidaac; 

SXaXSpOV XWV O, II 7IOIEIX63. 

'O^ioicdc; Sr| idle, erndvco Sel^ojiev, oxi oi A, A, E xal oi 
H, M, N, 9 'tE,f\z eiaiv dvdXoyov ev xw xou A 7tp6<; xov 
B Xoyw, xal exi oi E, S, Z xai oi 9, O, II, K zE,f\c, sioiv 
dvdXoyov sv ifi xou B upoc xov T Xoyco. xai eaxiv w<; 6 A 
npoc; xov B, ouxck 6 B npbc, xov E xal oi A, A, E apa xolc; 
E, S, Z ev xo auxo Xoyw elal xal exi oi H, M, N, 9 xolc 
9, O, II, K. xai eaxiv laov xo [Lev xwv A, A, E TtXrji9o<; xw 
xov E, S, Z TiXri'dei, xo 8s xov H, M, N, 9 xw xGv 9, O, 
II, K' Si' i'aou apa eaxiv <b<; ^xev 6 A npoc; xov E, ouxwc; 6 
E 7ip6<; xov Z, (b? 8e 6 H npoc xov 9, ouxgk 6 9 Ttpoc; xov 
K- oTiep eSei 8eic;ai. 



be (continuously) proportional [and this always happens 
with the extremes] . 

Let A, B, C be any multitude whatsoever of contin- 
uously proportional numbers, (such that) as A (is) to B, 
so B (is) to C. And let A, B, C make D, E, F (by) 
multiplying themselves, and let them make G, H, K (by) 
multiplying D, E, F. I say that D, E, F and G, H, K are 
continuously proportional. 
A' ' L 



O 



F >- 



O 
M 
N 
P 

Q 



H>- 



K 

For let A make L (by) multiplying B. And let A, B 
make M, N, respectively, (by) multiplying L. And, again, 
let B make O (by) multiplying C. And let B, C make P, 
Q, respectively, (by) multplying O. 

So, similarly to the above, we can show that D, L, 
E and G, M, N, H are continuously proportional in the 
ratio of A to B, and, further, (that) E, O, F and H, P, Q, 
K are continuously proportional in the ratio of B to C. 
And as A is to B, so B (is) to C. And thus D, L, E are in 
the same ratio as E, O, F, and, further, G, M, N, H (are 
in the same ratio) as H, P, Q, K. And the multitude of 
D, L, E is equal to the multitude of E, O, F, and that of 
G, M, N, H to that of H, P, Q, K. Thus, via equality, as 
D is to E, so E (is) to F, and as G (is) to H, so H (is) to 
K [Prop. 7.14]. (Which is) the very thing it was required 
to show. 



18'. 

°Edv xexpdywvoc; xsxpdyovov ^expfj, xai f) nXsupd xr)v 
TtXeupdv [isxprpei- xal eav f] TtXeupa xrjv nXeupdv (iexpfj, xal 
6 xexpdywvoc; xov xexpdywvov [letprpei. 

'Eaxwaav xexpdywvoi dpid^iol oi A, B, nXeupai 8e auxwv 
Eaxcoaav oi T, A, 6 8s A xov B ^sxpdxco - Xeyto, oxi xai 6 
T xov A ^expsT. 



Proposition 14 

If a square (number) measures a(nother) square 
(number) then the side (of the former) will also mea- 
sure the side (of the latter) . And if the side (of a square 
number) measures the side (of another square number) 
then the (former) square (number) will also measure the 
(latter) square (number). 

Let A and B be square numbers, and let C and D be 
their sides (respectively). And let A measure B. I say that 
C also measures D. 



241 



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ELEMENTS BOOK 8 



a i 1 r i 1 

B i ' A i 1 

Ei ' 

"O r yap tov A TioXXauXaaidaac; tov E raDidTW oi A, E, 
B dpa l^fje; dvdXoyov slaiv sv xG xou T Ttpoc; tov A Xoyco. 
xal £7td 01 A, E, B s£rj<; dvdXoyov eiaiv, xal ^expsl 6 A tov 
B, ^.STpsT dpa xal 6 A tov E. xa[ sgtiv cb<; 6 A 7tp6<; tov E, 
0UT6X 6 T npbc, tov A- UETpsI dpa xal 6 T tov A. 

ITdXiv 8r] 6 T tov A ^STpsrcM- Xsyw, oti xal 6 A tov B 
^tSTpsT. 

Tt5v yap auTGv xaTaaxsuaaiJevTOv ouoick 8ei£o^ev, 
oti oi A, E, B e^f]z dvdXoyov eiaiv £v iu tou T 7ip6<; tov A 
Xoyco. xal etcsl eaTiv &>c, 6 T npbz tov A, outoc 6 A 7ip6<; 
tov E, ^STpeT 8e 6 T tov A, ^STpeT dpa xal 6 A tov E. xal 
eiaiv ol A, E, B i^f]z dvdXoyov ^xeTpeT dpa xal 6 A tov B. 

'Edv dpa TSTpdycovot; TSTpdywvov ^CTpfj, xal f] TtXeupa 
t/]v TtXeupdv [LSTpfioei- xal edv f) nXeupd ttjv nXeupdv (j.expf], 
xal 6 TSTpdywvoc; tov TSTpdywvov [Lsjpfiaei- oTtep eBei 
8eTc;ai. 



A 1 ' O ' 

B 1 1 D 1 

Ei 1 

For let C make E (by) multiplying D. Thus, A, E, 
B are continuously proportional in the ratio of C to D 
[Prop. 8.11]. And since A, E, B are continuously pro- 
portional, and A measures B, A thus also measures E 
[Prop. 8.7]. And as A is to E, so C (is) to D. Thus, C 
also measures D [Def. 7.20] . 

So, again, let C measure D. I say that A also measures 

B. 

For similarly, with the same construction, we can 
show that A, E, B are continuously proportional in the 
ratio of C to D. And since as C is to D, so A (is) to E, 
and C measures D, A thus also measures E [Def. 7.20]. 
And A, E, B are continuously proportional. Thus, A also 
measures B. 

Thus, if a square (number) measures a(nother) square 
(number) then the side (of the former) will also measure 
the side (of the latter) . And if the side (of a square num- 
ber) measures the side (of another square number) then 
the (former) square (number) will also measure the (lat- 
ter) square (number). (Which is) the very thing it was 
required to show. 



is'. 

'Edv xu(3oc; dpiduoc; xupov dpid^tov [iSTpr], xal f) TtXeupa 
t/]v TiXsupdv \±sjpfioei- xal edv f) TtXeupa tt]v icXeupdv ^xexpf), 
xal 6 xupo<; tov xupov ^iCTprjaei. 

Kupog yap dpi%6c; 6 A xupov tov B ^CTpeiTM, xal tou 
y.sv A uXeupa eaTto 6 T, tou 8s B 6 A- Xeyco, oti 6 T tov 
A [iETpa. 

a 1 1 r 1 1 

B 1 1 A 1 1 

E 1 1 Bi ' 

Hi ' KJ 1 

Z 1 

'O r yap eauTov TtoXXa7tXaaidaa<; tov E tioicitco, 6 8s A 



Proposition 15 

If a cube number measures a(nother) cube number 
then the side (of the former) will also measure the side 
(of the latter) . And if the side (of a cube number) mea- 
sures the side (of another cube number) then the (for- 
mer) cube (number) will also measure the (latter) cube 
(number). 

For let the cube number A measure the cube (num- 
ber) B, and let C be the side of A, and D (the side) of B. 
I say that C measures D. 

A 1 1 C' 1 

B 1 1 D 1 

Ei 1 Hi 1 

Gi 1 Ki 1 

F 1 1 

For let C make E (by) multiplying itself. And let 



242 



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ELEMENTS BOOK 8 



eauxov TCoXXaTtXaaidaac; tov H ttoieixm, xal exi 6 T xov A 
TtoXXaitXaaidaag xov Z [ttoieixw], exdxepog 8e xaiv T, A xov 
Z TioAXamXaaidaac; exdxepov xwv 9, K Ttoidxw. cpavspov 
8r], oxi ol E, Z, H xod oi A, 0, K, B £^fj<; dvdXoyov eiaiv 
sv xG xou r Ttpoc; xov A Xoyw. xal etxel oi A, 9, K, B e^fjc; 
dvdXoyov eiaiv, xal ^texpeT 6 A xov B, ^icxpeT dpa xal xov 
9. xai eaxiv cbc 6 A Ttpoc; xov 9, oux«<; 6 T 7tp6<; xov A- 
^texpeT dpa xal 6 T xov A. 

AXXa 8rj ^lexpeixo 6 T xov A- Xeyw, oxi xal 6 A xov B 
^exprpei. 

Twv yap auxwv xaxaaxeuaadevxwv b\±oic)q 8f) 8ei^o^.ev, 
oxi oi A, 9, K, B e^fjc; dvdXoyov eiaiv ev xw xou V npoc, 
xov A X6ya>. xal etcei 6 T xov A ^expert, xai eaxiv (i><; 6 T 
7ip6<; xov A, ouxw<; 6 A Ttpoc xov 9, xal 6 A dpa xov 9 
^lexper waxe xal xov B ^icxpel 6 A- oicep e8ei BeT^ai. 

'Eav xexpdywvoc; dprd^ioc; xexpdywvov dpii9|ji6v y.r) 
(lexprj, ou8e f\ TtXeupd xr)v TtXeupdv |iexpr|aef xdv f) TtXeupd 
x/]v TiXeupdv \±r\ [icxpfj, ou8e 6 xexpdywvoc; xov xexpdyovov 
^iexpf|aei. 

a i 1 r i 1 

B ' 1 A i 1 

'Tiaxwaav xexpdywvoi dpid^ioi oi A, B, nXeupal 8e auxfiiv 
eaxwaav oi T, A, xal [ir] ^texpeixM 6 A xov B- Xeyw, oxi ou8e 
6 T xov A (iexpa. 

Ei yap \±£Tpei 6 T xov A, \LSTpr\oei xal 6 A xov B. ou 
^expel 8e 6 A xov B- ouSe dpa 6 T xov A ^expr|aei. 

Mr] ^texpeixM [Srj] TtdXiv 6 T xov A- Xeyw, oxi ouSe 6 A 
xov B [LSTpfioei. 

Ei yap ^texpeT 6 A xov B, ^iexpf|aei xal 6 T xov A. ou 
^texpeT 8e 6 T xov A- oOB' dpa 6 A xov B ^expfjaa- oitep 
e8el SeTc;ai. 



'Eav xupog dpid^icx; xupov dpid^iov y.r] ^exp/j, ou8s f) 
TiXeupd xrjv TtXeupdv [lexpiqaei- xdv f) TtXeupd xrjv TtXeupdv 
[ir] tisxpf), ouSe 6 xupoc; xov xu[3ov ^expiqaei. 



D make G (by) multiplying itself. And, further, [let] C 
[make] F (by) multiplying D, and let C, D make iJ, if, 
respectively, (by) multiplying F. So it is clear that E, F, 
G and A, H, K, B are continuously proportional in the 
ratio of C to D [Prop. 8.12]. And since A, H, K, B are 
continuously proportional, and A measures B, (A) thus 
also measures H [Prop. 8.7]. And as A is to H, so C (is) 
to D. Thus, C also measures D [Dei. 7.20]. 

And so let C measure D. I say that A will also mea- 
sure B. 

For similarly, with the same construction, we can 
show that A, H, K, B are continuously proportional in 
the ratio of C to D. And since C measures D, and as C is 
to D, so A (is) to H, A thus also measures H [Dei. 7.20]. 
Hence, A also measures B. (Which is) the very thing it 
was required to show. 

Proposition 16 

If a square number does not measure a(nother) 
square number then the side (of the former) will not 
measure the side (of the latter) either. And if the side (of 
a square number) does not measure the side (of another 
square number) then the (former) square (number) will 
not measure the (latter) square (number) either. 

A i ' ' C' ' 

B i 1 Di 1 

Let A and B be square numbers, and let C and D be 
their sides (respectively). And let A not measure B. I say 
that C does not measure D either. 

For if C measures D then A will also measure B 
[Prop. 8.14]. And A does not measure B. Thus, C will 
not measure D either. 

[So], again, let C not measure D. I say that A will not 
measure B either. 

For if A measures B then C will also measure D 
[Prop. 8.14]. And C does not measure D. Thus, A will 
not measure B either. (Which is) the very thing it was 
required to show. 

Proposition 17 

If a cube number does not measure a(nother) cube 
number then the side (of the former) will not measure the 
side (of the latter) either. And if the side (of a cube num- 
ber) does not measure the side (of another cube number) 
then the (former) cube (number) will not measure the 
(latter) cube (number) either. 



243 



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ELEMENTS BOOK 8 



Ai 1 r> 1 Ai 1 O 

B 1 Ai 1 Bi 1 D^ 



Kupog yap apvd\ibc, 6 A xupov dpidjiov xov B [if\ (ie- 
xpeixo, xoti xou [iev A TtXeupd eaxw 6 T, xou Be B 6 A- 
Xeyw, oxi 6 T xov A ou ^ETpVjaei. 

Ei yap [xexpeT 6 T xov A, xal 6 A xov B [iexpf|aei. ou 
[xexpeT Be 6 A xov B- ouB' dpa 6 T xov A [xexpeT. 

AXXd 8rj [if] uxxpeixw 6 T xov A- Xeyw, oxi ouBe 6 A 
xov B [iexpf|aei. 

Et yap o A xov B (jiexpeT, xal 6 T xov A [istprpei. ou 
[lejpsi 8e 6 T xov A- ou8' apa 6 A xov B {izjpfioei- oTtep 
e8ei 8eTe;ai. 



ir]'. 

Auo ojioicov eTtiTte8«v dpidjifiv elc; [izaoz dvdXoyov eaxiv 
dprd^ioc;- xal 6 enimboz Ttpoc; xov emTteSov BiTtXaaiova Xoyov 
exei f] by-oXoyoc, TtXeupd Ttpoc; xf)v o^ioXoyov TtXeupdv. 

A i 1 B i - 1 

T i 1 Ei 1 

A i 1 Z i 1 



For let the cube number A not measure the cube num- 
ber B. And let G be the side of A, and D (the side) of B. 
I say that G will not measure D. 

For if G measures D then A will also measure B 
[Prop. 8.15]. And A does not measure B. Thus, G does 
not measure D either. 

And so let G not measure D. I say that A will not 
measure B either. 

For if A measures B then G will also measure D 
[Prop. 8.15]. And G does not measure D. Thus, A will 
not measure B either. (Which is) the very thing it was 
required to show. 

Proposition 18 

There exists one number in mean proportion to two 
similar plane numbers. And (one) plane (number) has to 
the (other) plane (number) a squared^ ratio with respect 
to (that) a corresponding side (of the former has) to a 
corresponding side (of the latter). 

A i 1 B i 1 

Ci 1 E i— ' 

Di F i 



H 1 

Tiaxwaav Buo o^ioioi CTtiTteSoi dpid^tol oi A, B, xal xou 
[lev A TtXeupal eaxioaav oi T, A dpidjioi, xou Be B oi E, 
Z. xal CTtel o^ioioi CTtiTteBoi eiaiv oi dvdXoyov e^ovxec; xdc; 
TtXeupdc;, eaxiv apa wq 6 T Ttpoc; xov A, ouxwc; 6 E Ttpoc; 
xov Z. Xeyco ouv, oxi xwv A, B etc; \izaoc, dvdXoyov eaxiv 
dpii9|ji6c, xal 6 A Ttpoc; xov B BiTtXaaiova Xoyov e/ei f)Ttep 6 
T Ttpoc; xov E f) 6 A Ttpoc; xov Z, xouxeaxiv rjnep f\ o^ioXoyoc; 
TtXeupd Ttpoc; xr)v o^ioXoyov [TtXeupdv]. 

Kal eitel eaxiv tbc; 6 T Ttpoc; xov A, ouxcog 6 E Ttpoc; xov 
Z, evaXXdi; apa eaxiv &>c, 6 T Ttpoc; xov E, 6 A Ttpoc; xov Z. 
xal eitel sitiiteSoc; eaxiv 6 A, TtXeupal Be auxou oi T, A, 6 A 
apa xov r TtoXXaTtXaaidaac; xov A TteTtoirjxev. Bid xd auxa 
8r) xal 6 E xov Z TtoXXaTtXaaidaac; xov B TteTtoirjxev. 6 A 
8r) xov E TtoXXaTtXaaidaac; xov H Ttoieixw. xal eicel 6 A xov 
\±zv T TtoXXaTtXaaidaac xov A Tteitofyxev, xov Be E TtoXXa- 
TtXaaidaac; xov H TtCTtoirjxev, eaxiv apa w<; 6 T Ttpoc; xov E, 
ouxcoc; 6 A Ttpoc; xov H. dXX' (be; 6 T itpoc; xov E, [ouxioc;] 
6 A Ttpoc; xov Z- xal wc; apa 6 A Ttpoc; xov Z, ouxwc; 6 A 
Ttpoc; xov H. itdXiv, CTtel 6 E xov [lev A TtoXXaTtXaaidaac; xov 
H TteTto[/]xev, xov 8e Z TtoXXaTtXaaidaac; xov B TtCTtoirjxev, 
eaxiv apa u; 6 A Ttpoc; xov Z, ouxwc; 6 H Ttpoc; xov B. 
e8eix , dr) Be xal u; 6 A Ttpoc; xov Z, ouxwc; 6 A Ttpoc; xov 



Gi 1 

Let A and £? be two similar plane numbers. And let 
the numbers C, D be the sides of A, and E, F (the sides) 
of B. And since similar numbers are those having pro- 
portional sides [Def. 7.21], thus as C is to D, so E (is) to 
F. Therefore, I say that there exists one number in mean 
proportion to A and B, and that A has to B a squared 
ratio with respect to that C (has) to E, or D to F — that is 
to say, with respect to (that) a corresponding side (has) 
to a corresponding [side] . 

For since as C is to D, so E (is) to F, thus, alternately, 
as C is to E, so D (is) to F [Prop. 7.13]. And since A is 
plane, and C, D its sides, D has thus made A (by) mul- 
tiplying C. And so, for the same (reasons), E has made 
B (by) multiplying F. So let D make G (by) multiplying 
E. And since £> has made A (by) multiplying C, and has 
made G (by) multiplying E, thus as C is to E, so A (is) to 
G [Prop. 7.17]. But as C (is) to E, [so] L> (is) to F. And 
thus as D (is) to F, so A (is) to G. Again, since E has 
made G (by) multiplying D, and has made B (by) multi- 
plying F, thus as D is to F, so G (is) to B [Prop. 7.17]. 
And it was also shown that as D (is) to F, so A (is) to G. 
And thus as A (is) to G, so G (is) to B. Thus, A, G, i? are 



244 



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ELEMENTS BOOK 8 



EL xai cb? dpa 6 A npoc; xov H, ouxgk 6 H npoc; xov B. oi 
A, H, B apa zEjff, dvdXoyov eiaiv. xc5v A, B dpa dc; [ieaoc; 
dvdXoyov eaxiv dpi-djioc- 

Aeyco 8r], oxi xai 6 A npoc; xov B 8mXaaiova Xoyov 
exei f]nep f] b\±6koxoq nXeupa npoc; xrjv ojaoXoyov nXeupdv, 
xouxeaxiv r]nep 6 T npoc; xov Erjo A npoc; xov Z. etc! yap 
oi A, H, B E^fjc; dvdXoyov slaw, 6 A npoc; xov B SmXaaiova 
Xoyov £)(£i rjnep npoc; xov H. xai eaxiv wc; 6 A npoc; xov H, 
ouxox 6 xe F npoc; xov E xai 6 A npoc; xov Z. xai 6 A apa 
npoc; xov B SmXaaiova Xoyov i)(£i rjnep 6 T npoc; xov E rj 
6 A npoc; xov Z- onep e8ei Sdc^ai. 

t Literally, "double". 



continously proportional. Thus, there exists one number 
(namely, G) in mean proportion to A and _B. 

So I say that A also has to B a squared ratio with 
respect to (that) a corresponding side (has) to a corre- 
sponding side — that is to say, with respect to (that) G 
(has) to E, or D to F. For since A, G, B are continuously 
proportional, A has to B a squared ratio with respect to 
(that A has) to G [Prop. 5.9]. And as A is to G, so C (is) 
to E, and Z) to F. And thus A has to B a squared ratio 
with respect to (that) C (has) to E, or D to F. (Which 
is) the very thing it was required to show. 



10'. 

Auo opioiov axepewv dpid^icov 8uo [Leaoi dvdXoyov 
ejjminxouaiv dpid^ioi- xai 6 axepeoc; npoc; xov o^toiov axepeov 
xpmXaaiova Xoyov eyei f\Ksp f] ojioXoyoc; nXeupa npoc; xrjv 
o^toXoyov nXeupdv. 

A i — — ^ — i r ' 

A^ ' 
E' ' 

B i 1 Z i 1 

@i 1 

K 1 -h 

M i 1 Ni 1 

A 1 1 3 1 1 

"Eaxcoaav Suo o^ioioi axepeoi oi A, B, xai xou \xev A 
nXeupai eaxtoaav oi T, A, E, xou 8s B oi Z, H, 0. xai end 
S^ioioi axepeoi eiaiv oi dvdXoyov exovxeg xdc; nXeupdc;, eaxiv 
dpa <b<; [lev 6 T npoc; xov A, ouxwc; 6 Z npoc; xov H, «<; Be 
6 A npoc; xov E, ouxwc; 6 H npoc; xov 9. Xeyw, oxi iSv A, 
B Suo \ieaoi dvdXoyov ejininxouaiv dptd^toi, xai 6 A npoc; 
xov B xpmXaaiova Xoyov e)(ei fjnep 6 T npoc; xov Z xai 6 
A npoc; xov H xai exi 6 E npoc; xov 0. 

c O r yap xov A noXXanXaaidaac; xov K noieixw, 6 8e 
Z xov H noXXanXaaidaac; xov A noieixw. xai end oi T, 
A xoic; Z, H ev x5 auxfi Xoycp eiaiv, xai ex [ie\ x«v T, 
A eaxiv 6 K, ex 8e xGv Z, H 6 A, oi K, A [dpa] ojioioi 
enmeBoi eiaiv dpidjioi- x«v K, A dpa etc; ^icaoc; dvdXoyov 
eaxiv dprd^ioc;. eax« 6 M. 6 M dpa eaxiv 6 ex x«v A, 
Z, tbc; ev xfi npo xouxou -dewpiqijiaxi eSei/'dr]. xai end 6 
A xov [iev T noXXanXaaidaac; xov K nenoi/jxev, xov 8e Z 
noXXanXaaidaac; xov M nenoi/jxev, eaxiv dpa cbc; 6 T npoc; 
xov Z, ouxgjc; 6 K npoc; xov M. dXX' o; 6 K npoc; xov M, 
6 M npoc; xov A. oi K, M, A dpa ec^rjc; eiaiv dvdXoyov ev 



Proposition 19 

Two numbers fall (between) two similar solid num- 
bers in mean proportion. And a solid (number) has to 
a similar solid (number) a cubed^ ratio with respect to 
(that) a corresponding side (has) to a corresponding side. 

A i 1 C 1 

D 

Ei 1 

B ' ' F ' ' 

G ' 

Hi ' 

K ' 

Mi 1 Ni 1 

L i 1 Oi ' 

Let A and B be two similar solid numbers, and let 
G, D, E be the sides of A, and F, G, H (the sides) of 
B. And since similar solid (numbers) are those having 
proportional sides [Def. 7.21], thus as C is to D, so F 
(is) to G, and as D (is) to E, so G (is) to H . I say that 
two numbers fall (between) A and B in mean proportion, 
and (that) A has to B a cubed ratio with respect to (that) 
C (has) to F, and D to G, and, further, E to H. 

For let G make K (by) multiplying D, and let F make 
L (by) multiplying G. And since G, D are in the same 
ratio as F, G, and K is the (number created) from (mul- 
tiplying) G, D, and L the (number created) from (multi- 
plying) F, G, [thus] K and L are similar plane numbers 
[Def. 7.21]. Thus, there exits one number in mean pro- 
portion to K and L [Prop. 8.18]. Let it be M. Thus, M is 
the (number created) from (multiplying) D, F, as shown 
in the theorem before this (one) . And since D has made 
K (by) multiplying G, and has made M (by) multiplying 
F, thus as G is to F, so K (is) to M [Prop. 7.17]. But, as 



245 



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ELEMENTS BOOK 8 



xco xou r Tipoc xov Z Xoyio. xal CTiei eaxiv &>z b T Tipoc xov 
A, ouxtoc 6 Z Tipoc xov H, evaXXdc' dpa eaxiv cbc 6 T Tipoc 
xov Z, ouxwc 6 A Tipoc xov H. Bid xa auxa Bf] xal «c 6 A 
Tipoc xov H, ouxcoc 6 E Tipoc xov 0. oi K, M, A apa e<;rjc 
etcriv dvdXoyov ev xe xw xou T Tipoc xov Z Xoyw xal tu 
xoO A Tipoc xov H xal exi xfi xou E Tipoc xov 9. exaxepoc 
8r) xov E, O xov M TioXXaTiXaaidaac exdxepov xwv N, 5 
itoidxco. xal inel axepeoc eaxiv 6 A, TiXeupal Be auxou eiaiv 
oi r, A, E, 6 E dpa xov ex xwv T, A TioXXaTiXaaidaac xov 
A TieTtoi7)xev. 6 Be ex xwv T, A eaxiv 6 K - 6 E dpa xov K 
TioXXaTiXaaidaac xov A 7ieuo(r]xev. Bid xa auxa Br] xal 6 9 
xov A TioXXaTiXaaidaac xov B TieTio[r|Xev. xal eitel 6 E xov 
K TioXXaTiXaaidaac xov A TieTioirjXev, dXXd ^xfjv xal xov M 
TioXXaTiXaaidaac xov N TieTioirjXev, eaxiv dpa 6; o K Tipoc 
xov M, ouxwc 6 A Tipoc xov N. cbc Be 6 K Tipoc xov M, 
ouxcoc o xe r Tipoc xov Z xal 6 A Tipoc xov H xal exi 6 E 
Tipoc xov 0- xal cbc apa 6 T Tipoc xov Z xal 6 A Tipoc xov 
H xal 6 E Tipoc xov 6, ouxcoc 6 A Tipoc xov N. TidXiv, etc el 
exaxepoc xcov E, xov M TioXXaTiXaaidaac exdxepov xcov 
N, 5 7ien;oir]xev, eaxiv apa 6 E Tipoc xov 0, ouxcoc 6 N 
Tipoc xov S. dXX' cbc 6 E Tipoc xov 0, ouxcoc o xe T Tipoc xov 
Z xal 6 A Tipoc xov IE xal cbc dpa 6 T Tipoc xov Z xal 6 A 
Tipoc xov H xal 6 E Tipoc xov 0, ouxcoc 6 xe A Tipoc xov N 
xal 6 N Tipoc xov S. TidXiv, ercel 6 xov M TioXXaTiXaaidaac 
xov 5 TieTioirjxev, dXXd \ir\v xal xov A TioXXaTiXaaidaac xov 
B TiCTioirjxev, eaxiv dpa cbc 6 M Tipoc xov A, ouxcoc 6 S Tipoc 
xov B. dXX' d>c 6 M Tipoc xov A, ouxcoc o xe T Tipoc xov Z 
xal 6 A Tipoc xov H xal 6 E Tipoc xov 0. xal cbc dpa 6 T 
Tipoc xov Z xal 6 A Tipoc xov H xal 6 E Tipoc xov 0, ouxcoc 
ou (iovov 6 5 Tipoc xov B, dXXd xal 6 A Tipoc xov N xal 6 
N Tipoc xov S. oi A, N, S, B apa e^fjc eiaiv dvdXoyov ev 
xolc eipr)[ievoic xcov TiXeupcbv Xoyoic. 

Aeyco, 6xi xal 6 A Tipoc xov B xpiTiXaaiova Xoyov 
e/ei fjTtep f\ o^ioXoyoc TiXeupa Tipoc xfjv b\±6koyov TiXeupdv, 
xouxeaxiv rjTiep 6 r dpid^ioc Tipoc xov Z rj 6 A Tipoc xov 
H xal exi 6 E Tipoc xov 0. etieI yap xeaaapec dtpi'djjtol e^c 
dvdXoyov eiaiv oi A, N, S, B, 6 A apa Tipoc xov B xpi- 
TiXaaiova Xoyov sjsi fpzzp 6 A Tipoc xov N. dXX'' cbc 6 A 
Tipoc xov N, ouxwc eSeix^ 6 xe T Tipoc xov Z xal 6 A Tipoc 
xov H xal exi 6 E Tipoc xov 0. xal 6 A dpa Tipoc xov B 
xpiTiXaaiova Xoyov zyzi fjTiep f\ ojioXoyoc TiXeupa Tipoc xr)V 
o^ioXoyov TiXeupdv, xouxeaxiv fjTiep 6 T dpid^ioc Tipoc xov 
Z xal 6 A Tipoc xov H xal exi 6 E Tipoc xov 0- oTiep eBei 
BeT^ai. 

t Literally, "triple". 

X . 

'Edv Buo dpidfjicbv elc ^xeaoc dvdXoyov E^uuTtxfj dpnf)^6c, 
ojjioioi ctiiticBoi saovxai oi dtpi'djiof. 



K (is) to M, (so) M (is) to L. Thus, K, M, L are contin- 
uously proportional in the ratio of C to F. And since as 
C is to D, so F (is) to G, thus, alternately, as C is to F, so 
D (is) to G [Prop. 7.13]. And so, for the same (reasons), 
as D (is) to G, so E (is) to H . Thus, AT, M, L are contin- 
uously proportional in the ratio of C to F, and of D to G, 
and, further, of E to iJ. So let E, H make AT, O, respec- 
tively, (by) multiplying M. And since A is solid, and C, 

D, E are its sides, E has thus made A (by) multiplying 
the (number created) from (multiplying) C, D. And K 
is the (number created) from (multiplying) C, D. Thus, 
E has made A (by) multiplying AT. And so, for the same 
(reasons), H has made B (by) multiplying L. And since 
£ has made A (by) multiplying K, but has, in fact, also 
made A^ (by) multiplying M, thus as K is to M, so A (is) 
to N [Prop. 7.17]. And as K (is) to M, so C (is) to F, 
and D to G, and, further, E to H. And thus as C (is) to 
F, and D to G, and F to H, so A (is) to N. Again, since 

E, H have made N, O, respectively, (by) multiplying M, 
thus as E is to H, so A^ (is) to O [Prop. 7.18]. But, as 
E (is) to H, so G (is) to F, and L> to G. And thus as G 
(is) to F, and D to G, and F to H, so (is) A to N, and 
AT to O. Again, since H has made O (by) multiplying M, 
but has, in fact, also made B (by) multiplying L, thus as 
M (is) to L, so O (is) to B [Prop. 7.17]. But, as M (is) 
to i, so G (is) to F, and D to G, and F to H. And thus 
as G (is) to F, and D to G, and F to H, so not only (is) 
O to F, but also A to AT, and N to O. Thus, A, N, O, 
B are continuously proportional in the aforementioned 
ratios of the sides. 

So I say that A also has to B a cubed ratio with respect 
to (that) a corresponding side (has) to a corresponding 
side — that is to say, with respect to (that) the number G 
(has) to F, or D to G, and, further, F to H. For since A, 
N, O, B are four continuously proportional numbers, A 
thus has to B a cubed ratio with respect to (that) A (has) 
to A^ [Def. 5.10]. But, as A (is) to AT, so it was shown (is) 
G to F, and D to G, and, further, F to H. And thus A has 
to B a cubed ratio with respect to (that) a corresponding 
side (has) to a corresponding side — that is to say, with 
respect to (that) the number G (has) to F, and D to G, 
and, further, F to H. (Which is) the very thing it was 
required to show. 



Proposition 20 

If one number falls between two numbers in mean 
proportion then the numbers will be similar plane (num- 



246 



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ELEMENTS BOOK 8 



Auo ydp dpid^wv xwv A, B sic \ieaoz dvdXoyov 
EjiKiKTETW dpid|ji6c; 6 r- Xeyto, oxi oi A, B 6u.oioi £tutce;8o[ 
eiaiv dpid^ioi. 

A i ' A ' ' 

r ' 1 Z' ' 



H ' 

ElX^cpi^waav [yap] eXd)(iaxoi dpidfioi xfiv xov auxov 
Xoyov eywxcov A, -T 01 ^> ^' todxi? dpa 6 A xov A 
\LZTpel xdi 6 E xov T. oadxic; 8r) 6 A xov A ^.expeT, xoaauxai 
^iovd8ec; eaxwaav ev tw Z' 6 Z dpa xov A TioXXaTiXaaidaac; 
xov A TieTioir)xev. uoxe 6 A euiTieBoc; eaxiv, TiXeupai 8s 
auxou oi A, Z. TidXiv, excel oi A, E eXd)(iaxo[ eiai xwv xov 
auxov Xoyov exovxov xoTc; T, B, iadxic; dpa 6 A xov T [ls- 
xpeT xal 6 E xov B. oadxic; 5f] 6 E xov B ^lexpeT, xoaaDxai 
^.ovdSec; eaxwaav ev iu H. 6 E dpa xov B ^.expel xaxd xdc; 
ev xQ H [iovd8a<;- 6 H dpa xov E TioXXaTiXaaidaac; xov B 
TCTCoirjxev. 6 B dpa CTUTieSoc; laxi, TiXeupai 8e auxou eiaiv 
oi E, H. oi A, B dpa etutisBoi eiaiv dpid^toi. Xeyto 8rj, oti 
xai o(i.oioi. ETiei yap ° Z xov ^tev A TioXXaTiXaaidaac; xov 
A TceTioir]xev, xov Be E TioXXaTiXaaidaac; xov T TieTcoirjxev, 
eaxiv dpa cbc; 6 A Tipoc; xov E, ouxwc; 6 A Tipoc; xov T, 
xouxeaxiv 6 T Tipoc; xov B. TidXiv, enei 6 E exdxepov xwv Z, 
H TioXXaTiXaaidaac; xouc; T, B TiCTio[r)xev, eaxiv dpa 6; 6 Z 
Tipoc; xov H, ouxwc; 6 T Tipoc; xov B. cbc 8s 6 T Tipoc; xov B, 
ouxoc; 6 A Tipoc; xov E - xal foe, dpa 6 A Tipoc; xov E, ouxoc; 
6 Z Tipoc; xov H - xai evaXXac; (be; 6 A Tipoc; xov Z, ouxwc; 6 
E Tipoc; xov H. oi A, B dpa o^ioioi etxltxsSoi dpi'd^.oi eiaiv ai 
yap TiXeupai auxwv dvdXoyov eiaiv oTiep e5ei BeT^ai. 



xa'. 

'Edv 8uo dpid^icov 8uo jieaoi dvdXoyov e^iTUTixoaiv 
6ipvQ[Lol, o^ioioi axepeoi eiaiv oi dpid^ioi. 

Auo ydp dpid|i65v xwv A, B 8uo [ieaoi dvdXoyov 
e^miTixexwaav dpidfioi oi T, A- Xeyw, oxi oi A, B 6[ioioi 
axepeoi eiaiv. 

EiXrjcp'dtoaav ydp eXd)(iaxoi dpi'djioi xfiv xov auxov 
Xoyov e^ovxeov xolc; A, T, A xpeTc; oi E, Z, H- oi dpa dxpoi 
auxCSv oi E, H upcoxoi Tipoc; dXXrjXouc; eiaiv. xai CTiel tuv 
E, H eTc; ^.eaoc; dvdXoyov ejjiTiCTixoxev dpvd^ioc; 6 Z, oi E, 
H dpa dpidjio! o^xoioi CTUTie8oi eiaiv. eaxoaav ouv xou fiev 



bers). 

For let one number C fall between the two numbers A 
and B in mean proportion. I say that A and B are similar 
plane numbers. 

A i D 1 

Ci 1 F i 1 



G' ' 

[For] let the least numbers, D and E, having the 
same ratio as A and C have been taken [Prop. 7.33]. 
Thus, D measures A as many times as E (measures) C 
[Prop. 7.20]. So as many times as D measures A, so 
many units let there be in F. Thus, F has made A (by) 
multiplying D [Def. 7.15]. Hence, A is plane, and D, 
F (are) its sides. Again, since D and E are the least 
of those (numbers) having the same ratio as C and B, 
D thus measures C as many times as E (measures) B 
[Prop. 7.20]. So as many times as E measures B, so 
many units let there be in G. Thus, E measures B ac- 
cording to the units in G. Thus, G has made B (by) mul- 
tiplying E [Def. 7.15]. Thus, B is plane, and E, G are 
its sides. Thus, A and B are (both) plane numbers. So I 
say that (they are) also similar. For since F has made A 
(by) multiplying D, and has made C (by) multiplying E, 
thus as D is to E, so A (is) to C — that is to say, C to B 
[Prop. 7.17]. t Again, since E has made C, B (by) multi- 
plying F, G, respectively, thus as F is to G, so C (is) to 
B [Prop. 7.17]. And as C (is) to B, so D (is) to E. And 
thus as D (is) to E, so F (is) to G. And, alternately, as D 
(is) to F, so E (is) to G [Prop. 7.13]. Thus, A and B are 
similar plane numbers. For their sides are proportional 
[Def. 7.21]. (Which is) the very thing it was required to 
show. 

C. Furthermore, it is not necessary to show that D : E :: A : C, 

Proposition 21 

If two numbers fall between two numbers in mean 
proportion then the (latter) are similar solid (numbers) . 

For let the two numbers C and D fall between the two 
numbers A and B in mean proportion. I say that A and 
B are similar solid (numbers) . 

For let the three least numbers E, F, G having the 
same ratio as A, C, D have been taken [Prop. 8.2]. Thus, 
the outermost of them, E and G, are prime to one an- 
other [Prop. 8.3]. And since one number, F, has fallen 
(between) E and G in mean proportion, E and G are 



t This part of the proof is defective, since it is not demonstrated that F x E = 
because this is true by hypothesis. 



247 



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ELEMENTS BOOK 8 



E rcXeupai oi 9, K, xou 8e H oi A, M. cpavepov apa eaxlv 
ex xou Tcpo toutou, oti oi E, Z, H e^fjc; eiaiv dvdXoyov ev 
xe tu xou 9 Ttpoc; tov A Xoycp xod tc5 xou K rcpoi; tov M. 
xal excel oi E, Z, H eXd)(iaTo[ eiai xwv tov auxov Xoyov 
e)(6vTWv xou; A, T, A, xai saxiv laov to TiXfj'Oog tGv E, 
Z, H tw icX^^ei tGv A, T, A, 81' i'aou apa eotiv &>c, 6 E 
Ttpoc; tov H, ouTtog 6 A rcpoc; tov A. oi Se E, H rcpcoTOi, oi 
8e 7ip«Toi xal eXd/iaTOi, oi Se eXd)(ioToi (iCTpouai Toug tov 
auTov Xoyov exovTac; auTolc; iadxig o ts jjiei^cov tov jiei^ova 
xai 6 eXdaawv tov eXdaaova, toutsotiv o tc rjyou^ievoc 
tov r)you[ievov xai 6 erco^evoc; tov ejco^ievov iadxic; apa 
6 E tov A [iSTpsi xal 6 H tov A. oadxic; 8f] 6 E tov A 
[Lexpei, ToaauTai [iovd8e<; eaTwaav ev tw N. 6 N apa tov E 
TcoXXaicXaaidaac tov A tctco^xsv. 6 8e E eaTiv 6 ex xuv 
9, K- 6 N apa tov ex t5v 9, K TtoXXanXaaidaai; tov A 
iceTcofyxev. axepeoc; apa laxlv 6 A, xcXeupal 8e auToO eiaiv 
oi 9, K, N. rcdXiv, enel oi E, Z, H eXd/iaToi eiai tGv tov 
auTov Xoyov e/ovTiov to!<; T, A, B, iadxi? apa 6 E tov T 
^.ETpei xai 6 H tov B. oadxu; 8r) 6 E tov T [iCTpel, ToaauTai 
^.ovdBec; eaTwaav ev xfi S. 6 H apa tov B ^CTpeT xaTa Tag 
ev t& S [iovdSac;- 6 S apa tov H TroXXanXaaidaac; tov B 
Tterconqxev. 6 8s: H eaTiv 6 ex twv A, M' 6 S apa tov ex 
tGv A, M noXXanXaaidaat; tov B TteTio[r]xev. aTepeoc; apa 
eaTiv 6 B, rcXeupai Se auTou eiaiv oi A, M, £• oi A, B apa 
cnxpeoi eiaiv. 



A i 1 ©i 1 

r ' ' K' ' 

A 1 ' N' ' 

B ' ' 

Ei ' A 1 1 

Z ' ' Mi ' 

H 1 1 H 1 1 

Aeyw [8r]], oti xai o^toioi. creel yap oi N, H tov E tcoX- 
XarcXaaidaavTec tou<; A, T Tcercoiiqxaaiv, eaTiv apa cbc; 6 N 
rcpoc; tov S, 6 A rcpoc; tov T, toutcotiv 6 E rcpoc; tov Z. 
dXX' cbc; 6 E rcpoc; tov Z, 6 9 rcpoc; tov A xal 6 K rcpoc; tov 
M- xai (b? apa 6 9 rcpoc; tov A, outwc; 6 K rcpoc; tov M xai 
6 N Ttpoc tov H. xai eiaiv oi piev 9, K, N rcXeupai tou A, 



thus similar plane numbers [Prop. 8.20]. Therefore, let 
H, K be the sides of E, and L, M (the sides) of G. Thus, 
it is clear from the (proposition) before this (one) that E, 
F, G are continuously proportional in the ratio of H to 
L, and of K to M. And since E, F, G are the least (num- 
bers) having the same ratio as A, C, D, and the multitude 
of E, F, G is equal to the multitude of A, C, D, thus, 
via equality, as E is to G, so A (is) to D [Prop. 7.14]. 
And E and G (are) prime (to one another), and prime 
(numbers) are also the least (of those numbers having 
the same ratio as them) [Prop. 7.21], and the least (num- 
bers) measure those (numbers) having the same ratio as 
them an equal number of times, the greater (measuring) 
the greater, and the lesser the lesser — that is to say, the 
leading (measuring) the leading, and the following the 
following [Prop. 7.20]. Thus, E measures A the same 
number of times as G (measures) D. So as many times as 
E measures A, so many units let there be in N. Thus, N 
has made A (by) multiplying E [Dei. 7.15]. And E is the 
(number created) from (multiplying) H and K. Thus, 
has made A (by) multiplying the (number created) from 
(multiplying) H and K. Thus, A is solid, and its sides are 
H , K, N. Again, since E, F, G are the least (numbers) 
having the same ratio as C, D, B, thus E measures C the 
same number of times as G (measures) B [Prop. 7.20]. 
So as many times as E measures C, so many units let 
there be in O. Thus, G measures B according to the units 
in O. Thus, O has made B (by) multiplying G. And 
G is the (number created) from (multiplying) L and M. 
Thus, O has made B (by) multiplying the (number cre- 
ated) from (multiplying) L and M. Thus, B is solid, and 
its sides are L, M, O. Thus, A and B are (both) solid. 



Ah 
O 



L h 



Mh 



Gh 



Oh 



[So] I say that (they are) also similar. For since N, O 
have made A, C (by) multiplying E, thus as A^ is to O, so 
A (is) to C — that is to say, E to F [Prop. 7.18]. But, as 
E (is) to F, so H (is) to L, and K to M. And thus as H 
(is) to L, so K (is) to M, and N to O. And H, K, N are 
the sides of A, and L, M, O the sides of B. Thus, A and 



248 



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ELEMENTS BOOK 8 



oi 8e S, A, M TtXeupai xou B. oi A, B dpa dpi-d^oi o^oioi 
axepeoi etcnv onep eSsi 8eT<;ai. 

t The Greek text has "O, L, M", which is obviously a mistake. 

x(3'. 

'Edv xpelc dpi%oi e^fjc; dvdXoyov ffioiv, 6 Ss Tipcoxog 
xexpdytovot; fj, xai 6 xpixo<; xexpdywvoc; Eaxai. 

A i 1 

B i 1 

r ' ' 

"Eaxwaav xpeu; dpid^ioi £E,ffc dvdXoyov oi A, B, T, 6 8e 
Tipwxoi; 6 A xExpaycovoc saxw Xsyw, oxi xal 6 xpixoc; 6 T 
xexpdycovoc; saxiv. 

'Etc! yap xwv A, r zic, \itaoc, dvdXoyov saxiv dpid^icx; 
6 B, oi A, r dpa o^ioioi eiiLKeSoi eiaiv. xexpdytdvoc 8s: 6 A- 
xexpdywvo<; dpa xai 6 E ojtep e8ei 8eTc;ai. 

xy'. 

'Eav xeaaapec; dpid^oi e^rjc; dvdXoyov Sow, 6 Se upaixog 
xupog fj, xai 6 xexapxoc xupoc eaxai. 

A' ■ 

B i ' 

r i 1 

A | 1 

'Eaxwaav xeaaape<; dpid^ioi e<;fj<; dvdXoyov oi A, B, T, 
A, 6 8s A x6po<; eaxw Xeyco, oxi xai 6 A x6po<; eaxiv. 

'Ercei yap xwv A, A 8uo \±eooi dvdXoyov eiaiv dpidjjioi 
oi B, T, oi A, A dpa opioioi eiai axepeoi dpi%oi. xupoc; 8s 
6 A- x6po<; dpa xai 6 A- onep eBei SeT^ai. 



x5'. 

'Eav 8uo dpiiD^toi Ttpoc; dXXf]Xou<; Xoyov e/waiv, ov 
xexpdywvoc; dpnSuoc; xpoc; xexpdyovov dpid^tov, 6 8e 
Ttpwxoc; xexpdyiovoc; fj, xai 6 Seuxepoc; xexpdytovoc; eaxai. 

A i ' i r i ' 

B i 1 A i 1 

Auo yap dpi%ol oi A, B Ttpoc; dXXiqXouc; Xoyov 



B are similar solid numbers [Def. 7.21]. (Which is) the 
very thing it was required to show. 



Proposition 22 

If three numbers are continuously proportional, and 
the first is square, then the third will also be square. 

A' 1 

B' 1 

C' ' 

Let A, B, C be three continuously proportional num- 
bers, and let the first A be square. I say that the third C 
is also square. 

For since one number, B, is in mean proportion to 
A and C, A and C are thus similar plane (numbers) 
[Prop. 8.20]. And A is square. Thus, C is also square 
[Def. 7.21]. (Which is) the very thing it was required to 
show. 

Proposition 23 

If four numbers are continuously proportional, and 
the first is cube, then the fourth will also be cube. 

A' ' 

B' ' 

C' 1 

D 1 

Let A, B, C, D be four continuously proportional 
numbers, and let A be cube. I say that D is also cube. 

For since two numbers, B and C, are in mean propor- 
tion to A and D, A and D are thus similar solid numbers 
[Prop. 8.21]. And A (is) cube. Thus, D (is) also cube 
[Def. 7.21]. (Which is) the very thing it was required to 
show. 

Proposition 24 

If two numbers have to one another the ratio which 
a square number (has) to a(nother) square number, and 
the first is square, then the second will also be square. 

A i 1 C' ' 

B i 1 D 1 

For let two numbers, A and B, have to one another 



249 



STOIXEIQN rf. 



ELEMENTS BOOK 8 



e/exGjaav, ov xexpdycovoc; dpid^oc; 6 T 7tp6<; xexpdyovov 
dpi-Ojiov xov A, 6 8e A xexpdycovoc eaxto - Xeyw, on xal 6 
B xexpdycovoc eaxiv. 

'Eitei yap oi T, A xexpdycovoi eiaiv, oi T, A dpa ouoioi 
eraueSoi eiaiv. xcov T, A dpa etc ^jieaoc dvdXoyov e^tTUKxei 
dpid^oc. xa( eaxiv cbc 6 T Ttpoc xov A, 6 A Ttpoc xov B- 
xal xov A, B dpa eTc; jieaoc dvdXoyov e^jutmxei dpid^ioc;. xa[ 
eaxiv 6 A xexpdycovoc;- xal 6 B dpa xexpdycovoc eaxiv oTtep 
e8ei SeTcai. 



the ratio which the square number C (has) to the square 
number D. And let A be square. I say that B is also 
square. 

For since C and D are square, C and D are thus sim- 
ilar plane (numbers). Thus, one number falls (between) 
C and D in mean proportion [Prop. 8.18]. And as C is 
to D, (so) A (is) to B. Thus, one number also falls (be- 
tween) A and B in mean proportion [Prop. 8.8]. And A 
is square. Thus, B is also square [Prop. 8.22]. (Which is) 
the very thing it was required to show 



xs'. 

'Edv 8uo dprd^iol npoc dXXf|Xo\Jc Xoyov e/coaiv, ov 
xuf3o<; dpi^oc rcpoc xu[3ov dpiduov, 6 8e itpcoxoc xu[3oc; rj, 
xal 6 Beuxepoc xupoc eaxai. 

a i 1 r ' ' 

E' ' 

Z' 1 

B i 1 A i 1 

Auo yap apidjjioi oi A, B Ttpoc dXXiqXouc Xoyov 
e^excoaav, ov xupoc dpidjioc 6 T Ttpoc xu^ov dpidjiov xov 
A, xupoc 8e eaxco 6 A- Xeyco [8r|], oxi xal 6 B xupoc eaxiv. 

'End yap oi T, A xu[3oi eiaiv, oi T, A ouoioi axe- 
peoi eiaiv xcov T, A dpa 8uo ueaoi dvdXoyov euTUTtxouaiv 
dprduoi. oaoi 8e eic xouc T, A uexacu xaxd xo auvexec 
dvdXoyov euTUTtxouaiv, xoaouxoi xal eic xouc xov auxov 
Xoyov e)(ovxac auxolc' coaxe xal xcov A, B 8uo ueaoi 
dvdXoyov euTUTtxouaiv apiduoi. euTtiTtxexcoaav oi E, Z. eitei 
ouv xeaaapec dpi-d^ol oi A, E, Z, B ei;rjc dvdXoyov eiaiv, 
xai eaxi xupoc 6 A, xupoc dpa xal 6 B- oitep eBei 8eTc;ai. 



Proposition 25 

If two numbers have to one another the ratio which 
a cube number (has) to a(nother) cube number, and the 
first is cube, then the second will also be cube. 
A i ' C' 1 

Ei ' 

F ' ' 

B i 1 D 1 

For let two numbers, A and B, have to one another 
the ratio which the cube number C (has) to the cube 
number D. And let A be cube. [So] I say that B is also 
cube. 

For since C and D are cube (numbers), C and D are 
(thus) similar solid (numbers). Thus, two numbers fall 
(between) C and D in mean proportion [Prop. 8.19]. 
And as many (numbers) as fall in between C and D in 
continued proportion, so many also (fall) in (between) 
those (numbers) having the same ratio as them (in con- 
tinued proportion) [Prop. 8.8]. And hence two numbers 
fall (between) A and B in mean proportion. Let E and F 
(so) fall. Therefore, since the four numbers A, E, F, B 
are continuously proportional, and A is cube, B (is) thus 
also cube [Prop. 8.23]. (Which is) the very thing it was 
required to show. 



xt'. Proposition 26 

Oi ouoioi eTUTieSoi dpii9uoi Ttpoc dXXiqXouc Xoyov exou- Similar plane numbers have to one another the ratio 
aiv, ov xexpdycovoc dpi$u6c Ttpoc xexpdycovov dpnf)uov. which (some) square number (has) to a(nother) square 

number. 

A' 1 A 1 A' 1 Di 1 



r> 1 E' ' C' 1 

B ' 1 Z 1 B 1 F ^ 



"Eaxcoaav ouoioi ctuttcSoi apvduoi oi A, B- Xeyco, oxi Let A and B be similar plane numbers. I say that A 
6 A Ttpoc xov B Xoyov e/ei, ov xexpdycovoc dpiiSuoc Ttpoc has to B the ratio which (some) square number (has) to 



250 



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ELEMENTS BOOK 8 



xexpdywvov dpiduov. 

Tkel ydp oi A, B ojioioi eiuTteSoi eiaiv, t«v A, B dpa 
sic; \isaoc, dvdXoyov cjituttxci dpi%6c;. e^ranxexco xdi eaxto 6 
r, xdi eiXr|cpiL>Gjaav eXd)(iaxoi dpid^ioi x65v xov auxov Xoyov 
e/ovxcov xou; A, r, B oi A, E, Z- oi dpa dxpoi auxcov oi 
A, Z xexpdycovoi daw. xdi enei eaxiv (be 6 A rcpoc; xov Z, 
ouxcoc; 6 A Ttpoc; xov B, xai eiaiv oi A, Z xexpdywvoi, 6 A 
dpa rcpoc; xov B Xoyov exei, ° v te^pdycovoc dpid^io? -rcpoc; 
xexpdycovov dpid^tov orcep eBei BeT^ai. 



Oi o^toioi axepeoi dpid^toi 7ip6<; dXXr|Xou<; Xoyov e)(ou- 
olv, ov xupoc; dpid^toc; 7tp6<; xupov dpid^tov. 

A 1 ' Ei ' 

r ' ' z ' ' 

A 1 1 H' ' 

B i 1 ©| 1 

'Eaxcoaav o^toioi axepeoi dpid^toi oi A, B' Xeyw, oxi 6 
A npbc, xov B Xoyov e/ei, ov xufioc, dpid^oc; Kpo<; xupov 
dpid^iov. 

Tkei ydp oi A, B o^ioioi axepeoi eiaiv, xwv A, B dpa 8uo 
^xeaoi dvdXoyov e^KiKxouaiv dpid^oi. e^XTiiTixexwaav oi T, 
A, xai eiXf](pTf)waav eXd)(iaxoi dpid^toi xwv xov auxov Xoyov 
exovxwv xolc A, T, A, B laoi auxoTg xo nXfj'doc; oi E, Z, H, 
9- oi dpa dxpoi auxwv oi E, xOpoi eiaiv. xai eaxiv cbc 6 
E 7tp6<; xov 6, ouxw<; 6 A 7tp6<; xov B- xai 6 A dpa npbc, 
xov B Xoyov exei, Sv x6po<; dpi'djioc; npoc, xupov dpi%6v 
oTiep e8ei SeT^ai. 



a(nother) square number. 

For since A and B are similar plane numbers, one 
number thus falls (between) A and B in mean propor- 
tion [Prop. 8.18]. Let it (so) fall, and let it be C. And 
let the least numbers, D, E, F, having the same ratio 
as A, C, B have been taken [Prop. 8.2]. The outermost 
of them, D and F, are thus square [Prop. 8.2 corr.]. And 
since as D is to F, so A (is) to B, and D and F are square, 
A thus has to B the ratio which (some) square number 
(has) to a(nother) square number. (Which is) the very 
thing it was required to show. 

Proposition 27 

Similar solid numbers have to one another the ratio 
which (some) cube number (has) to a(nother) cube num- 
ber. 

A i 1 E i 1 

Ci 1 F i 1 

Di 1 G' 1 

B 1 Hi 1 

Let A and B be similar solid numbers. I say that A 
has to B the ratio which (some) cube number (has) to 
a(nother) cube number. 

For since A and B are similar solid (numbers), two 
numbers thus fall (between) A and B in mean proportion 
[Prop. 8.19]. Let C and D have (so) fallen. And let the 
least numbers, E, F, G, H, having the same ratio as A, 
C, D, B, (and) equal in multitude to them, have been 
taken [Prop. 8.2]. Thus, the outermost of them, E and 
H, are cube [Prop. 8.2 corr.]. And as E is to H, so A (is) 
to B. And thus A has to B the ratio which (some) cube 
number (has) to a(nother) cube number. (Which is) the 
very thing it was required to show. 



251 



252 



ELEMENTS BOOK 9 



Applications of Number 77ieory| 



tThe propositions contained in Books 7-9 are generally attributed to the school of Pythagoras. 



253 



ETOIXEIfiN fl'. 



ELEMENTS BOOK 9 



a . 

'Edv 860 6[ioioi eiuTieSoi dpidjiol TtoXXanXaaidaavxec; 
dXXr]Xouc; noiwai xiva, 6 yevo^ievoc; xexpdycovoc; eaxai. 

A' 1 

B ' ' 

r 1 1 

A' ' 

"Eaxwaav 860 8(ioioi eiuneSoi dpi%ol 01 A, B, xal 6 
A xov B TtoXXaTtXaaidaac; xov T Ttoieixw Xeyw, oxi 6 T 
xexpdycovo? eaxiv. 

c yap A eauxov rcoXXaTtXaaidaa? xov A uoieixw. 6 A 
dpa xexpdytovo? eaxiv. inel ouv 6 A eauxov [lev TtoXXa- 
TtXaaidaac; xov A Tte7toir)xev, xov 8s B rcoXXaTtXaaidaac; xov 
r TiSKOi/jxev, eaxiv dpa 6 A 7tp6<; xov B, oux«<; 6 A 
7ipo<; xov T. xal etiei 01 A, B o^ioioi CTUTteSoi eiaiv dpid^ol, 
x£Sv A, B dpa elc, [leooc, dvdXoyov IjiitiinEi dpid^to?. eav 8e 
860 dpid^iGv [icxa^u xaxa xo auvexe? dvdXoyov e^miKxoaiv 
dpi'd^oi, oaoi ei? auxou? e^TUTtxouai, xoaouxoi xal el? xou? 
xov auxov Xoyov exovxag- &axe xal xwv A, T eT? [leaoq, 
dvdXoyov e^tiitixei dpid^o?. xal eaxi xexpdywvo? 6 A- 
xexpdyovo? dpa xal 6 E oitep e§ei 8eTc;ai. 



P'- 

'Eav 860 dpid^iol 7toXXaTtXaaidaavxe<; dXXr]Xou<; itoicoai 
xsxpdywvov, 0^.0101 ekikeBoi eiaiv dpid^.01. 

A' 1 

B 1 1 

r 1 1 

A' ■ 

TEaxwaav 860 dpi-duol 01 A, B, xal 6 A xov B KoXXa- 
TtXaaidaac; xexpdycovov xov T noislxw Xeyw, oxi oi A, B 
ojioioi eraTieSoi eiaiv dpidjioi. 

yap A eauxov TroXXanXaaidaac; xov A Tioieixor 6 A 
dpa xexpdytovo? eaxiv. xal end 6 A eauxov (lev TtoXXa- 
TtXaaidaac; xov A Kenoi/jxev, xov 8e B TioXXanXaoidaat; xov 
r nenonqxev, eaxiv dpa 6 A Ttpoc; xov B, 6 A Ttpoc; xov 
T. xal stcI 6 A xexpdywvog eaxiv, dXXd xal 6 T, oi A, T 
dpa ofioioi eniTieSoi eiaiv. xfiv A, T dpa si? [isaog dvdXoyov 



Proposition 1 

If two similar plane numbers make some (number by) 
multiplying one another then the created (number) will 
be square. 

A' ' 

B 1 

C' ' 

D 1 

Let A and B be two similar plane numbers, and let A 
make C (by) multiplying B. I say that C is square. 

For let A make D (by) multiplying itself. D is thus 
square. Therefore, since A has made D (by) multiply- 
ing itself, and has made C (by) multiplying B, thus as A 
is to B, so D (is) to C [Prop. 7.17]. And since A and 
B are similar plane numbers, one number thus falls (be- 
tween) A and B in mean proportion [Prop. 8.18]. And if 
(some) numbers fall between two numbers in continued 
proportion then, as many (numbers) as fall in (between) 
them (in continued proportion), so many also (fall) in 
(between numbers) having the same ratio (as them in 
continued proportion) [Prop. 8.8]. And hence one num- 
ber falls (between) D and C in mean proportion. And D 
is square. Thus, C (is) also square [Prop. 8.22]. (Which 
is) the very thing it was required to show. 

Proposition 2 

If two numbers make a square (number by) multiply- 
ing one another then they are similar plane numbers. 

A' ' 

B 1 

C' ' 

D 1 

Let A and B be two numbers, and let A make the 
square (number) C (by) multiplying B. I say that A and 
B are similar plane numbers. 

For let A make D (by) multiplying itself. Thus, D is 
square. And since A has made D (by) multiplying itself, 
and has made C (by) multiplying B, thus as A is to B, so 
D (is) to C [Prop. 7.17]. And since D is square, and C 
(is) also, D and C are thus similar plane numbers. Thus, 
one (number) falls (between) D and C in mean propor- 



254 



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ELEMENTS BOOK 9 



ejitutctei. xa[ eaxiv cbc; 6 A Ttpoc; xov T, ouxcoc; 6 A Ttpoc; xov 
B- xod xfiv A, B apa elc; \±£ooz dvdXoyov e^miuxei. edv 8s 
8uo dpi%cov sic, [Leaoc, dvdXoyov e^tutitt], o^toioi stutieSoi 
eiaiv [oi] dpid^ioi- oi apa A, B o^toioi eiaiv eiuTieBoi- oitep 
eSei 8eTc;ai. 



Y'- 

'Edv xupoc; dpn^oc; eauxov TtoXXarcXaaidaac; uoifj xiva, 
6 yevojievoc; xupoc; eaxai. 

A i 1 

B i 1 

r i 1 

A i ' 

Kupoc; yap dpidjjidc; 6 A eauxov KoXXaTtXaaidaac; xov B 
itoieixco- Xeyo, oxi 6 B xupoc; eaxiv. 

ElXf](p , dcL) yap xou A TtXeupd 6 T, xal 6 T eauxov KoXXa- 
TtXaaidaac; xov A noieixco. cpavepov 8Vj eaxiv, oxi 6 T xov A 
TioXXaiiXaaidaac; xov A TteTCofyxev. xal cuei 6 T eauxov tioX- 
XaTtXaaidaac; xov A Ti£7io[r)X£v, 6 T apa xov A ^.expel xaxa 
xdc; ev auxo \xov6&u.c,. dXXd ^irjv xal f\ [iovdc xov T ^.expeT 
xaxa xdc; ev auxfi uovdBac;- eaxiv apa cbc; f] \xovac, Tipoc; xov 
r, 6 r Ttpoc; xov A. TtdXiv, CTiel 6 T xov A TioXXaTtXaaidaac; 
xov A 7i£iio[r]xev, 6 A apa xov A jiexpeT xaxa xdc ev xw T 
^iovd8ac;. ^texpel 8e xal f) ^tova<; xov T xaxa xdc; ev auxfi 
^tovd8ac;' eaxiv apa 6c; f] ^lovdc; Tipoc; xov T, 6 A Tipoc; xov 
A. dXX' wc; f) [iovdc; Tipoc; xov T, 6 T Tipoc; xov A- xal dbc; 
apa f) jiovdc; Tipoc; xov T, ouxcoc; 6 T Tipoc; xov A xal 6 A 
Tipoc; xov A. xrjc; apa jiovdSoc; xal xou A dpid^iou 8uo ^.eaoi 
dvdXoyov xaxa xo auvexec; ejiTCCTixoxaaiv dpiiJ^ol oi T, A. 
TtdXiv, etc el 6 A eauxov TioXXaTtXaaidaac; xov B TieTtoirjxev, 
6 A apa xov B (jiexpeT xaxa xdc; ev auxw ^ovd8ac;' [icxpel 8e 
xal f) jiovdc; xov A xaxa xdc; ev auxCS jiovdBac;- eaxiv apa (be; 
f] ^.ovdc; Tipoc; xov A, 6 A Tipoc; xov B. xfjc; 8e [lovdBoc; xal 
xou A 8uo ^.eaoi dvdXoyov ejiiceuxoxaaiv dpi'd^oi' xal x£>v 
A, B apa Buo \ieaoi dvdXoyov ejiiceaouvxai dpi'd^ioi. edv 8e 
8uo dpi-d^wv 8uo ^teaoi dvdXoyov e^miiixwaiv, 6 8e TipGxoc; 
xupoc; fj, xal 6 Beuxepoc; xupoc; eaxai. xa( eaxiv 6 A xupoc 
xal 6 B apa xupoc; eaxiv oTtep eSei Belial. 

5'. 

'Edv xupoc; dpid^ioc; xupov dpid^iov TioXXaTtXaaidaac; 
Ttoifj xiva, 6 yev6[ievoc; xupoc; eaxai. 



tion [Prop. 8.18]. And as D is to C, so A (is) to B. Thus, 
one (number) also falls (between) A and B in mean pro- 
portion [Prop. 8.8]. And if one (number) falls (between) 
two numbers in mean proportion then [the] numbers are 
similar plane (numbers) [Prop. 8.20]. Thus, A and B are 
similar plane (numbers). (Which is) the very thing it was 
required to show. 

Proposition 3 

If a cube number makes some (number by) multiply- 
ing itself then the created (number) will be cube. 

A' 1 

Bi 1 

Ci ' 

D 1 

For let the cube number A make B (by) multiplying 
itself. I say that B is cube. 

For let the side C of A have been taken. And let C 
make D by multiplying itself. So it is clear that C has 
made A (by) multiplying D. And since C has made D 
(by) multiplying itself, C thus measures D according to 
the units in it [Def. 7.15]. But, in fact, a unit also mea- 
sures C according to the units in it [Def. 7.20] . Thus, as 
a unit is to C, so C (is) to D. Again, since C has made A 
(by) multiplying D, D thus measures A according to the 
units in C. And a unit also measures C according to the 
units in it. Thus, as a unit is to C, so D (is) to A. But, 
as a unit (is) to C, so C (is) to D. And thus as a unit (is) 
to C, so C (is) to D, and D to A. Thus, two numbers, C 
and D, have fallen (between) a unit and the number A 
in continued mean proportion. Again, since A has made 
B (by) multiplying itself, A thus measures B according 
to the units in it. And a unit also measures A according 
to the units in it. Thus, as a unit is to A, so A (is) to B. 
And two numbers have fallen (between) a unit and A in 
mean proportion. Thus two numbers will also fall (be- 
tween) A and B in mean proportion [Prop. 8.8]. And if 
two (numbers) fall (between) two numbers in mean pro- 
portion, and the first (number) is cube, then the second 
will also be cube [Prop. 8.23]. And A is cube. Thus, B 
is also cube. (Which is) the very thing it was required to 
show. 

Proposition 4 

If a cube number makes some (number by) multiply- 
ing a(nother) cube number then the created (number) 



255 



ETOIXEIfiN fl'. 



ELEMENTS BOOK 9 



A' ' 

B i 1 

r ' ' 

a i 1 

Kupoc; yap dpi'd^oc; 6 A xupov dpid^iov xov B TtoXXa- 
TiXaoidaac; xov T tcoieitw Xeyto, oxi 6 T xupoc; eaxiv. 

c O yap A eauxov TtoXXarcXaaidaac; xov A Ttoieixw 6 
A dpa xupoc; eaxiv. xal ercel 6 A eauxov uev TtoXXa- 
TtXaaidaac; xov A TteTtoirjxev, xov 8s B rcoXXaTiXaaidaac; xov 
r K£7io[r)X£v, eaxiv dpa ox; 6 A rcpoc; xov B, ouxoc; 6 A Ttpoc; 
xov T. xal ETtsi oi A, B xupoi eiaiv, o^ioioi axepeoi eiaiv oi A, 
B. x«v A, B dpa 80o [leaoi dvdXoyov ejiTunxouaiv dpi%o£- 
Saxe xal x«v A, r Buo [isaoi dvdXoyov ejjmeaouvxai 
dpii9|jioL. xa[ eaxi xupoc; 6 A- xupoc; dpa xal 6 T- onep 
eSei 8e"i£ai. 

e'. 

°Eav xupog dpi-d^oc; dpi%6v xiva TioXXaTtXaaidaac; xupov 
Tioifj, xal 6 TtoXXaTiXaaiatydeic; xupoc; eaxai. 

A i 1 

B i 1 

r i 1 

A' ■ 

Kupoc; yap dpi'djioc; 6 A apiduov xiva xov B TtoXXa- 
TtXaaidaac; xupov xov T noieixw Xeyw, oxi 6 B xupoc; eaxiv. 

c yap A eauxov TtoXXaitXaaidaac; xov A Ttoieixco - xupoc; 
dpa saxiv 6 A. xal ercel 6 A eauxov ^tev TCoXXanXaaidaac; xov 
A 7ieTio(r]xev, xov 8e B TtoXXaitXaaidaac; xov T TiCTtoirjxev, 
eaxiv dpa (be; 6 A Ttpoc; xov B, 6 A Ttpoc; xov 1\ xal excel 
oi A, T xupoi eioiv, ojioioi axepeoi eioiv. xfiv A, T dpa 
Suo [ieaoi dvdXoyov eu-iciicxouaiv dpid^oi. xai eaxiv wc; 6 A 
jepoe; xov r, ouxwc; 6 A Tipoc; xov B- xai xwv A, B dpa 8uo 
[icooi dvdXoyov ejiTUTixouaiv dpidjioi. xai eaxi xupoc; 6 A- 
xupoc; dpa eaxi xal 6 B- onep e8ei 8el^ai. 



'Eav dpidfjioc; eauxov TcoXXaTiXaaidaac; xupov noifj, xal 



will be cube. 

A' ' 

B' 1 

Ci 1 

Di 1 

For let the cube number A make C (by) multiplying 
the cube number B. I say that C is cube. 

For let A make D (by) multiplying itself. Thus, D is 
cube [Prop. 9.3]. And since A has made D (by) multi- 
plying itself, and has made C (by) multiplying B, thus 
as A is to B, so D (is) to C [Prop. 7.17]. And since A 
and B are cube, A and B are similar solid (numbers). 
Thus, two numbers fall (between) A and B in mean pro- 
portion [Prop. 8.19]. Hence, two numbers will also fall 
(between) D and C in mean proportion [Prop. 8.8]. And 
D is cube. Thus, C (is) also cube [Prop. 8.23]. (Which 
is) the very thing it was required to show. 

Proposition 5 

If a cube number makes a(nother) cube number (by) 
multiplying some (number) then the (number) multi- 
plied will also be cube. 

A' 1 

B' 1 

O ' 

D' 1 

For let the cube number A make the cube (number) 
C (by) multiplying some number B. I say that B is cube. 

For let A make D (by) multiplying itself. D is thus 
cube [Prop. 9.3]. And since A has made D (by) multiply- 
ing itself, and has made C (by) multiplying B, thus as A 
is to B, so D (is) to C [Prop. 7.17]. And since D and C 
are (both) cube, they are similar solid (numbers) . Thus, 
two numbers fall (between) D and C in mean proportion 
[Prop. 8.19]. And as D is to C, so A (is) to B. Thus, 
two numbers also fall (between) A and B in mean pro- 
portion [Prop. 8.8]. And A is cube. Thus, B is also cube 
[Prop. 8.23]. (Which is) the very thing it was required to 
show. 

Proposition 6 
If a number makes a cube (number by) multiplying 



256 



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ELEMENTS BOOK 9 



auxog xu[3o<; eoxai. 

A' 1 

B i 1 

r i 1 

'Api'd(i6<; yap 6 A eauTov TtoXXaTtXaaidaac; xupov tov B 
tioieitw Xeyw, oxi xal 6 A xupoc; eaTiv. 

c yap A tov B TioXXaTiXaaidaag tov T Ttoierao. enel ouv 
6 A eauTov ^iev 7ioXXaTtXaaidaa<; tov B TtCTtoirjxev, tov Be: 
B TioXXaTtXaaidaac; tov V Tienoirjxsv, 6 T dpa xupoc; eaTiv. 
xal ettsI 6 A eauTov TtoXXaTtXaaidaac; tov B kstioi/jxev, 6 
A dpa tov B ^lETpel xaTa Tag ev auTfi jiovdSac;. (iCTpel Be 
xal f) ^iovd<; tov A xaTa Tag sv auTfi (jiovaBag. eaTiv dpa 
cbc f] jiovac; 7ip6<; tov A, outw<; 6 A 7tp6<; tov B. xal ensl 
6 A tov B TtoXXaTtXaaidaac; tov T TteTTOirjxsv, 6 B dpa tov 
r (jLexpeT xaTa Tag ev to A ^ovd8a<;. ^.eTpel 8e xal rj ^ovdc 
tov A xaTa Ta<; ev auTW ^iovd8a<;. eaTiv dpa w<; f] [Lovaz 
7ipo<; tov A, outw<; 6 B Ttpoc; tov T. dXX' d><; f] [lovolq npbz 
tov A, outoc 6 A rcpoc; tov B- xal (i><; dpa 6 A upoc tov B, 
6 B npoz tov T. xal eitel ol B, T xupoi eiaiv, ojioioi cnxpeoi 
eiaiv. twv B, T dpa 8uo [isaoi dvdXoyov eiaiv dpi'd^oi. xai 
eaTiv 6 B Ttp6<; tov T, 6 A Ttpoc; tov B. xal twv A, B 
dpa 8uo [Leaoi dvdXoyov eiaiv dpid^ioi. xa[ eaTiv xupoc; 6 
B- xupoc; dpa eotI xal 6 A- ouep e8ei Sel^ai. 

c 

'Eav auvdeToc; dpid^toc; dpidjiov Tiva TioXXajcXaaidaac; 
icoifj Tiva, 6 yevojievoc; oxepebq eaxca. 

A' 1 

B i 1 

r i 1 

A i ' 

Ei 1 

SuvdsTog yap dpid^ioc; 6 A dpid^ov Tiva tov B TtoXXa- 
jcXaaidaac; tov T uoieiTW Xeyw, oti 6 T aTepeoc; eaTiv. 

'End yap 6 A auvdeToc; eaTiv, Otto dpi^ou tivoc; (ie- 
Tp/)i9r]aeTai. |ieTpe[ad« utto tou A, xal oadxic; 6 A tov A 
(iCTpel, ToaaOTai (iovd8e<; eaTwaav sv t£) E. insi ouv 6 A 
tov A (iETpel xaTa Tag ev tG E |iovd8a<;, 6 E dpa tov A 
TtoXXaTtXaaidaac; tov A TteitoiTjxev. xal eitel 6 A tov B ttoX- 
XanXaaidaac; tov T Tieno[r)xev, 6 Se A eaTiv 6 ex to>v A, E, 
6 dpa ex t«v A, E tov B TtoXXaitXaaidaac; tov T nenoi/jxev. 
6 T dpa aTepeoc; eaTiv, icXeupal 5e auToO eiaiv oi A, E, B- 



itself then it itself will also be cube. 

A' 1 

B 1 

C' 1 

For let the number A make the cube (number) B (by) 
multiplying itself. I say that A is also cube. 

For let A make C (by) multiplying B. Therefore, since 
A has made B (by) multiplying itself, and has made C 
(by) multiplying B, C is thus cube. And since A has made 
B (by) multiplying itself, A thus measures B according to 
the units in (A). And a unit also measures A according 
to the units in it. Thus, as a unit is to A, so A (is) to 
B. And since A has made C (by) multiplying B, B thus 
measures C according to the units in A. And a unit also 
measures A according to the units in it. Thus, as a unit is 
to A, so B (is) to C. But, as a unit (is) to A, so A (is) to 
B. And thus as A (is) to B, (so) B (is) to C. And since B 
and C are cube, they are similar solid (numbers) . Thus, 
there exist two numbers in mean proportion (between) 
B and C [Prop. 8.19]. And as B is to C, (so) A (is) to B. 
Thus, there also exist two numbers in mean proportion 
(between) A and B [Prop. 8.8]. And B is cube. Thus, A 
is also cube [Prop. 8.23]. (Which is) the very thing it was 
required to show. 

Proposition 7 

If a composite number makes some (number by) mul- 
tiplying some (other) number then the created (number) 
will be solid. 

A' ' 

B 1 

C' 1 

Di 1 

Ei ' 

For let the composite number A make C (by) multi- 
plying some number B. I say that C is solid. 

For since A is a composite (number), it will be mea- 
sured by some number. Let it be measured by D. And, 
as many times as D measures A, so many units let there 
be in E. Therefore, since D measures A according to 
the units in E, E has thus made A (by) multiplying D 
[Def. 7.15]. And since A has made C (by) multiplying B, 
and A is the (number created) from (multiplying) D, E, 
the (number created) from (multiplying) D, E has thus 



257 



ETOIXEIfiN fl'. 



ELEMENTS BOOK 9 



onep e8ei SeT^ai. 



'Edv and jiovdBoc; ottoooiouv dpi'd^iol e^fjc; dvdXoyov 
CSaiv, 6 jiev xpixoc dfto xfjt; ^lovdBoc; xexpdycovoc; eaxai 
xal oi eva SiaXeiTtovxec;, 6 8e xexapxoc; xupoc; xal oi 
86o SiaXeinovxec; Ttdvxec;, 6 8s epSojioc; xupoc; djia xal 
xexpdycovoc; xal oi rcevxe SiaXeiTtovxec;. 



A' 1 

B i 1 

r i 1 

A | 1 

Ei 1 

Z' 1 

"Eaxwaav duo ^tovdSoc; otioooiouv dpid^ioi e^fjc; dvdXoy- 
ov oi A, B, r, A, E, Z- Xeyto, oxi 6 \±ev xpixoc; duo 
xrjc ^ovdSoc; 6 B xexpdytovoc; eaxi xal oi eva SiaXeiTtovxec; 
Ttdvxec;, 6 8s xexapxoc; 6 T xupoc; xal oi 66o SiaXeiTtovxec; 
Ttdvxec;, 6 8s epBojjioc; 6 Z xupoc; d^ia xal xexpdycovoc; xal oi 
Ttevxe SiaXeiTtovxec; Ttdvxec;. 

'EtcI yap ectxiv toe; f] jiovdc; Ttpoc; xov A, ouxmc; 6 A 
Ttpoc; xov B, iadxic; dpa f) ^lovdc; xov A dpidjiov [icxpel xal 
6 A xov B. f] Se jiovdc; xov A dpnSjiov jiexpeT xaxa xdc; 
ev auxfi [iovdBac;- xal 6 A dpa xov B [iexpel xaxa xdc; ev 
xfi A [lovdSac;. 6 A dpa eauxov TtoXXaTtXaaidaac; xov B 
7ienoiT)X£V xexpdycovoc; dpa eaxlv 6 B. xal eTtel oi B, T, A 
e<;rjc; dvdXoyov eiaiv, 6 8s B xexpdycovoc; eaxiv, xal 6 A dpa 
xexpdycovoc; eaxiv. Sid xd auxd 8r) xal 6 Z xexpdycovoc; 
eaxiv. ojioicoc; 8/) 8ei^o[i£v, oxi xal oi eva SiaXeiTtovxec; 
Ttdvxec; xexpdycovoi eiaiv. Xeyco 6r], oxi xal 6 xexapxoc; and 
xrjg ^xovdSoc; 6 T xupoc eaxi xal oi 66o SiaXeiTtovxec; Ttdvxec;. 
inzi yap eoxiv cbc; f] [iomolq Ttpoc; xov A, ouxcoc; 6 B Ttpoc; xov 
r, iadxic; dpa f\ jiovac; xov A dpid^tov (icxpel xal 6 B xov T. f) 
Se fiovac; xov A dpid^iov ^.expel xaxa xdc; ev xcp A jiovdSac;- 
xal 6 B dpa xov T ^texpel xaxa xdc; ev iu A ^tovdSac;- 6 A 
dpa xov B TtoXXaTtXaaidaac; xov T TteTtonqxev. eTtel ouv 6 
A eauxov [ikv TtoXXaTtXaaidaac; xov B TteTtoirjxev, xov Se B 
TtoXXaTtXaaidaac; xov T TteTtoirjxev, xupoc; dpa eoxiv 6 T. xal 
eTtel oi T, A, E, Z sZ,f\c, dvdXoyov eiaiv, 6 Se T xupoc; eaxiv, 



made C (by) multiplying B. Thus, C is solid, and its sides 
are D, E, B. (Which is) the very thing it was required to 
show. 

Proposition 8 

If any multitude whatsoever of numbers is continu- 
ously proportional, (starting) from a unit, then the third 
from the unit will be square, and (all) those (numbers 
after that) which leave an interval of one (number), and 
the fourth (will be) cube, and all those (numbers after 
that) which leave an interval of two (numbers), and the 
seventh (will be) both cube and square, and (all) those 
(numbers after that) which leave an interval of five (num- 
bers). 

A' ' 

B 1 

C' ' 

D 1 

Ei ' 

F i ' 

Let any multitude whatsoever of numbers, A, B, C, 
D, E, F, be continuously proportional, (starting) from 
a unit. I say that the third from the unit, B, is square, 
and all those (numbers after that) which leave an inter- 
val of one (number). And the fourth (from the unit), C, 
(is) cube, and all those (numbers after that) which leave 
an interval of two (numbers) . And the seventh (from the 
unit), F, (is) both cube and square, and all those (num- 
bers after that) which leave an interval of five (numbers) . 

For since as the unit is to A, so A (is) to B, the unit 
thus measures the number A the same number of times 
as A (measures) B [Def. 7.20]. And the unit measures 
the number A according to the units in it. Thus, A also 
measures B according to the units in A. A has thus made 
B (by) multiplying itself [Def. 7.15]. Thus, B is square. 
And since B, C, D are continuously proportional, and B 
is square, D is thus also square [Prop. 8.22]. So, for the 
same (reasons), F is also square. So, similarly, we can 
also show that all those (numbers after that) which leave 
an interval of one (number) are square. So I also say that 
the fourth (number) from the unit, C, is cube, and all 
those (numbers after that) which leave an interval of two 
(numbers). For since as the unit is to A, so B (is) to C, 
the unit thus measures the number A the same number 
of times that B (measures) C. And the unit measures the 



258 



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ELEMENTS BOOK 9 



xal 6 Z dpa xupo<; eaxiv. £8e[)cdT] Be: xal xexpdycovoc;' 6 dpa 
ipBo^iot; duo xfjc [iovd8o<; xupoc; xe eaxi xal xexpdycovo<;. 
o^lolwc; 5f) Bei^o|jiev, oxi xal oi Ttevxe 8iaXemovxe<; 7tdvxe<; 
xupoi xe eiai xal xexpdywvor orcep eBei BeT^ai. 



0'. 

'Eav aTio jiovdBoc; oitoaoiouv e^rjc; xaxa xo auvexec 
dpi%ol dvdXoyov waiv, 6 Be ^exd xiqv ^xovdBa xexpdycovoc 
fj, xal oi Xomol ndvxec; xexpdycovoi eaovxai. xal eav 6 (icxa 
xr]v (jiovdSa xupo<; fj, xal oi XoitcoI Ttdvxec; xupoi eaovxai. 

A' 1 

B i 1 

r i 1 

A | ' 

E' 1 

Z' ' 

"Eaxwaav cmo ^iovdBo<; 'tEjff, dvdXoyov 6aoiBr)Ttoxouv 
dipi'd^ol oi A, B, T, A, E, Z, 6 Be ^texa xf)v ^xovdBa 
6 A xexpdywvoc; eaxw Xey«, oxi xal oi Xomol udvxsc 
xexpdytovoi eaovxai. 

"Oxi fiev ouv 6 xpixoc duo xfj? jiovdBoc 6 B xexpdy«vo<; 
eaxi xal oi eva SiaTiXemovxet; Ttdvxec;, SeBeixxar Xeyw [8yj], 
oxi xal oi Xomol navxec; xexpdycovoi eiaiv. Emel yap oi A, 
B, T e^fj<; dvdXoyov eiaiv, xai eaxiv 6 A xexpdytovo<;, xal 6 
T [apa] xexpdywvoc; eaxiv. TtdXiv, tnei [xal] oi B, T, A e^rj<; 
dvdXoyov eiaiv, xai eaxiv 6 B xexpdywvo^, xai 6 A [apa] 
xexpdywvo<; eaxiv. 6^ioi«<; Br) Bei^ouev, oxi xal oi Xoircoi 
iravxec xexpdyovoi eiaiv. 

AXXa 8r] eaxco 6 A xupoc Xeyco, oxi xal oi Xomol Ttdvxec 
xupoi eiaiv. 

"Oxi [iev ouv 6 xexapxoc; duo xfjc; ^ovdBoc; 6 T xupo<; eaxi 
xal oi 86o SiaXemovxec 7tdvxe<;, BeBeixxar Xeyco [Br]], oxi xai 
oi Xomol 7tdvxe<; xupoi eiaiv. end yap eaxiv &>z f) [iova<; npbz 
xov A, ouxw<; 6 A Ttpoc; xov B, iadxic; apa r\ (iovac xov A 
[iexpel xai 6 A xov B. f) Be ^ovdc; xov A (iexpeT xaxa xac; ev 



number A according to the units in A. And thus B mea- 
sures C according to the units in A. A has thus made C 
(by) multiplying B. Therefore, since A has made B (by) 
multiplying itself, and has made C (by) multiplying B, C 
is thus cube. And since C, D, E, F are continuously pro- 
portional, and C is cube, F is thus also cube [Prop. 8.23]. 
And it was also shown (to be) square. Thus, the seventh 
(number) from the unit is (both) cube and square. So, 
similarly, we can show that all those (numbers after that) 
which leave an interval of five (numbers) are (both) cube 
and square. (Which is) the very thing it was required to 
show. 

Proposition 9 

If any multitude whatsoever of numbers is continu- 
ously proportional, (starting) from a unit, and the (num- 
ber) after the unit is square, then all the remaining (num- 
bers) will also be square. And if the (number) after the 
unit is cube, then all the remaining (numbers) will also 
be cube. 

A' ' 

B 1 

C' ' 

D 1 

Ei ' 

F i ' 

Let any multitude whatsoever of numbers, A, B, C, 
D, E, F, be continuously proportional, (starting) from a 
unit. And let the (number) after the unit, A, be square. I 
say that all the remaining (numbers) will also be square. 

In fact, it has (already) been shown that the third 
(number) from the unit, B, is square, and all those (num- 
bers after that) which leave an interval of one (number) 
[Prop. 9.8]. [So] I say that all the remaining (num- 
bers) are also square. For since A, B, C are continu- 
ously proportional, and A (is) square, C is [thus] also 
square [Prop. 8.22]. Again, since B, C, D are [also] con- 
tinuously proportional, and B is square, D is [thus] also 
square [Prop. 8.22]. So, similarly, we can show that all 
the remaining (numbers) are also square. 

And so let A be cube. I say that all the remaining 
(numbers) are also cube. 

In fact, it has (already) been shown that the fourth 
(number) from the unit, C, is cube, and all those (num- 
bers after that) which leave an interval of two (numbers) 



259 



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ELEMENTS BOOK 9 



auxo) jiovdBac;' xal 6 A dpa xov B [iexpel xaxd xag ev aux£> 
(iovdSac 6 A apa eauxov TroXXanXaaidaac; tov B tcseoi/jxev. 
xat eaxiv 6 A x6po<;. edv 8e x6po<; dpiduoc; eauxov TtoXXa- 
TtXaaidaac; Ttoirj xiva, 6 yevo^ievo<; x6po<; eaxiv xal 6 B dpa 
xOpoc eaxiv. xal six si xeaaapec; dpid^tol oi A, B, T, A iZ,f\c, 
dvdXoyov eiaiv, xa[ eaxiv 6 A xupog, xal 6 A apa xupoc 
eaxiv. Bid xd auxd Sr] xal 6 E xupoc eaxiv, xal 6^toi«<; oi 
XoitcoI Kdvxsc xupoi eiaiv oTtep eBei BeT^ai. 



[Prop. 9.8]. [So] I say that all the remaining (numbers) 
are also cube. For since as the unit is to A, so A (is) to 
B, the unit thus measures A the same number of times 
as A (measures) B. And the unit measures A according 
to the units in it. Thus, A also measures B according to 
the units in (A) . A has thus made B (by) multiplying it- 
self. And A is cube. And if a cube number makes some 
(number by) multiplying itself then the created (number) 
is cube [Prop. 9.3]. Thus, B is also cube. And since the 
four numbers A, B, C, D are continuously proportional, 
and A is cube, D is thus also cube [Prop. 8.23]. So, for 
the same (reasons), E is also cube, and, similarly, all the 
remaining (numbers) are cube. (Which is) the very thing 
it was required to show. 



i . 

'Edv and ^ovd§o<; otcoctoioOv dpiif)[iol [ec;fj<;] dvdXoyov 
Gaiv, 6 Be ^.exd x/]v jiovdBa \±r\ fj xexpdytovoc;, ou5' aXXoc; 
ouSelc; xexpdycovoc eaxai x w P^ TO ° tpttou duo xrjg jiovdSoc; 
xal xwv eva BiaXemovxwv jcdvxwv. xal edv 6 jiexd x/]v 
uovdSa xupoc [ir] fj, o05e aXXo<; ouBelc; xOpoc eaxai x w P^ 
xoO xexdpxou dno xfjc; uovdSoc xal x«v 8uo SiaXeinovxov 
ndvxwv. 

A' 1 

B i 1 

r i 1 

A | ' 

E' 1 

Z' 1 

TCaxcoaav duo ^tovdSoc; e^fj<; dvdXoyov 6aoi8r)n;oxouv 
dpid^tol oi A, B, T, A, E, Z, 6 uexd x/]v ^tovdSa 6 A [iy] 
eaxw xexpdywvoc Xey«, oxi ouSe akXoc, o05elc; xexpdycovoc; 
eaxai x w P^ TO ° tpfrou dico xr]<; uovdBoc; [xal xwv eva 8ia- 
Xeikovxwv] . 

EE yap 5uvax6v, eaxw 6 T xexpdywvoc;. eaxi 8e xal 6 
B xexpdywvoc oi B, T apa 7ipo<; dXXr|Xou<; Xoyov exou- 
aiv, ov xexpdytovoc dpidjjioc; Ttpoc xexpdywvov dpid^iov. xai 
eaxiv &>z 6 B Ttpoc; xov T, 6 A Ttpoc; xov B' oi A, B dpa 
Ttpoc; dXXrjXouc; Xoyov exouaiv, ° v xexpdytovoc; dpid^ioc; Ttpoc; 
xexpdyiovov dpi%6v uoie oi A, B ouoioi CTtiTteSoi eiaiv. 
xai eaxi xexpdytovoc; 6 B- xexpdycovoc; apa eaxi xal 6 A- 
OTtep oux unexeixo. oux apa 6 T xexpdytovoc; eaxiv. o^ioitoc; 
8f) Bei^o^tev, oxi oOS' dXXoc; ouSelc; xexpdyovoc eaxi x w P^ 



Proposition 10 

If any multitude whatsoever of numbers is [continu- 
ously] proportional, (starting) from a unit, and the (num- 
ber) after the unit is not square, then no other (number) 
will be square either, apart from the third from the unit, 
and all those (numbers after that) which leave an inter- 
val of one (number) . And if the (number) after the unit 
is not cube, then no other (number) will be cube either, 
apart from the fourth from the unit, and all those (num- 
bers after that) which leave an interval of two (numbers). 

A' ' 

B 1 

C' ' 

D 1 

Ei 1 

F i ' 

Let any multitude whatsoever of numbers, A, B, C, 
D, E, F, be continuously proportional, (starting) from 
a unit. And let the (number) after the unit, A, not be 
square. I say that no other (number) will be square ei- 
ther, apart from the third from the unit [and (all) those 
(numbers after that) which leave an interval of one (num- 
ber)]. 

For, if possible, let C be square. And B is also 
square [Prop. 9.8]. Thus, B and C have to one another 
(the) ratio which (some) square number (has) to (some 
other) square number. And as B is to C, (so) A (is) 
to B. Thus, A and B have to one another (the) ratio 
which (some) square number has to (some other) square 
number. Hence, A and B are similar plane (numbers) 



260 



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ELEMENTS BOOK 9 



xou xplxou duo xfjc; ^xovdBoc; xal xfiv eva BiaXeiTxovxojv. 

AXXd Bf] (jltq eaxw 6 A xupoc;. Xeyw, oxi o08' dXXoc; 
ouBelc; xupoc; eaxai x w P^ TO ° texdpxou duo xfjc; jiovdBoc; 
xal xov Buo BiaXemovxwv. 

El yap Buvaxov, saxw 6 A xupoc;. eaxi Be: xal 6 T xupoc 
xexapxoc; yap feaxiv duo xfjc; [LovaSoq,. xal eaxiv cbc; 6 T Tipoc; 
xov A, 6 B npbc, xov E xal 6 B apa Tipoc; xov T Xoyov exei, 
6v xOpoi; Tipoc; xupov. xal eaxiv 6 T xupoc xal 6 B apa 
xOpoc eaxlv. xal inzi saxiv dbc; f] ^iovd<; Tipoc; xov A, 6 A Tipoc; 
xov B, f\ Be [iovac; xov A ^.expei xaxa xac; ev aux<2 jiovdSac 
xal 6 A apa xov B ^.expeT xaxa xac; ev auxw ^.ovdBac 6 A 
apa eauxov TioXXaTiXaaidaac; xupov xov B TieTiolrjxev. eav 
Be dpid^toc; eauxov TioXXaTiXaaidaac; xupov Tioifj, xal auxoc 
xupoc; eaxai. xupoc; apa xal 6 A- ouep oux UTioxeixai. oux 
apa 6 A xupoc; eaxlv. o^tolwc; 8f] Bel^o^tev, oxi ouB' aXXoc; 
ouBelc; xupoc; eaxi X W P^ x °u xexdpxou arco xfjc; ^tovdBoc; xal 
iSv Buo SiaXemovxcov ouep eSsri Belial. 



ia'. 

'Eav dno jiovdSoc; ottoooiouv dpi'd^ol e^fjc; dvdXoyov 
Sow, 6 eXdxxwv xov jiel^ova ^icxpeT xaxa xiva xcov unap/ovx- 
cov ev xolc; dvdXoyov apti^ou;. 

A' 1 

B 

r 

A 

E 

'Eaxwaav diio (jiovdSoc; xfjc; A oTioaoiouv dpid^iol ec;fjc; 
dvdXoyov ol B, T, A, E- Xey«, oxi x<2v B, T, A, E 6 
eXdxiaxoc; 6 B xov E ^lexpeT xaxa xiva xwv T, A. 

'End yap eaxiv wc; f) A ^.ovac; Tipoc; xov B, ouxwc; 6 A 
Tipoc; xov E, ladxic; apa f] A [lovac; xov B dpid^tov ^.expeT 
xal 6 A xov E- evaXXdJ; apa ladxic; f) A ^lovac; xov A ^texpeT 
xal 6 B xov E. f) Be A jiovdc; xov A jiexpe'i xaxa xac; ev 



[Prop. 8.26]. And B is square. Thus, A is also square. 
The very opposite thing was assumed. C is thus not 
square. So, similarly, we can show that no other (number 
is) square either, apart from the third from the unit, and 
(all) those (numbers after that) which leave an interval 
of one (number). 

And so let A not be cube. I say that no other (num- 
ber) will be cube either, apart from the fourth from the 
unit, and (all) those (numbers after that) which leave an 
interval of two (numbers) . 

For, if possible, let D be cube. And C is also cube 
[Prop. 9.8]. For it is the fourth (number) from the unit. 
And as C is to D, (so) B (is) to C. And B thus has to 
C the ratio which (some) cube (number has) to (some 
other) cube (number). And C is cube. Thus, B is also 
cube [Props. 7.13, 8.25]. And since as the unit is to 
A, (so) A (is) to B, and the unit measures A accord- 
ing to the units in it, A thus also measures B according 
to the units in (A). Thus, A has made the cube (num- 
ber) B (by) multiplying itself. And if a number makes a 
cube (number by) multiplying itself then it itself will be 
cube [Prop. 9.6]. Thus, A (is) also cube. The very oppo- 
site thing was assumed. Thus, D is not cube. So, simi- 
larly, we can show that no other (number) is cube either, 
apart from the fourth from the unit, and (all) those (num- 
bers after that) which leave an interval of two (numbers). 
(Which is) the very thing it was required to show. 

Proposition 11 

If any multitude whatsoever of numbers is continu- 
ously proportional, (starting) from a unit, then a lesser 
(number) measures a greater according to some existing 
(number) among the proportional numbers. 

A' ' 

B 1 

O 1 

D 1 



Let any multitude whatsoever of numbers, B, C, D, 
E, be continuously proportional, (starting) from the unit 
A. I say that, for B, C, D, E, the least (number), B, 
measures E according to some (one) of C, D. 

For since as the unit A is to B, so D (is) to E, the 
unit A thus measures the number B the same number of 
times as D (measures) E. Thus, alternately, the unit A 



261 



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ELEMENTS BOOK 9 



auxo) (iovdBac xal 6 B apa xov E [iexpei xaxd xac; ev xfi 
A ^iovd8a<;- <2>axe 6 eXdaaov 6 B xov ^.d^ova xov E ^e- 
xpeT xaxd xiva dpid^iov xwv UTtapxovxtov ev tou; dvdXoyov 
dpid^iou;. 



I16pia[Jia. 

Kal cpavepov, oxi rjv £)(£i xd^iv 6 (iexpwv aTto ^ovd8o<;, 
x/]v auxiqv e)(ei xal 6 xocd' 6v [iexpel arco xou (iexpou^evou 
em xo Tipo auxou. onep eSsi Sei^ai. 



LP'. 

'Edv dno [iovd8o<; onoaoiouv dpid^iol e^fjc dvdXoyov 
Goiv, ucp' oacov dv 6 eayoLxoz uptoxwv dpidjiaiv ^.expfjxat,, 
utio xwv auxwv xal 6 Tiapa xfjv ^ovd8a ^.£xpr]Tf)r]aexai. 

A i ' Ei 1 

B i 1 Z i 1 

r i 1 Hi 1 

A i 1 ©i 1 

'Eaxwaav duo ^iovdBo<; OTCoaoiBrjTtoxouv dpid^tol dvdXoy- 
ov oi A, B, r, A' Xsyw, oxi ucp' oouv dv 6 A upcoxcov 
dpn3jji£5v jisxpfjxai, utio xwv auxcov xal 6 A [lexprydrpsxa!.. 

Mexpelo'dw yap 6 A utio xivoc Ttpwxou dpii&jiou xou E' 
Xeyw, oxi 6 E xov A [isxpel. \±r\ yap' xai eoxiv 6 E Tcpwxog, 
aTca<; 8e Ttp£>xo<; dpi'djioc; Ttpoc; auavxa, ov \Lr\ [LSTpei, Tipwxoc; 
eaxiv oi E, A apa Ttpoxoi Ttpoc dXXiqXouc eiaiv. xal stceI 
6 E xov A [isxpei, ^.expeixw auxov xaxd xov Z' 6 E apa 
xov Z TtoXXaTtXaaidaag xov A TtSTtoirjXEv. TtdXiv, snel 6 A 
xov A [isxpsl xaxd xdc; ev xw T jiovdBac;, 6 A apa xov T 
TtoXXaTiXaaidaac xov A TieTcoirjXEv. dXXa \±r\v xal 6 E xov Z 
TtoXXaitXaaidaac; xov A TC£Tio[/]xev 6 apa tx xGv A, T Xaoz 
sraxl iw ex x£>v E, Z. saxiv apa 6? 6 A Ttpoc; xov E, 6 Z 
Tcp6<; xov T. oi 8e A, E TtpQxoi, oi 8e Tcpwxoi xal sXd)(iaxoi, 
oi 8s eXd)(iaxoi ^lexpouai xouc; xov auxov Xoyov s/ovxag 
iadxu; o xe f]you\L£voz xov f]you^ievov xal 6 stc6^svo<; xov 
STc6[ievov (jtexpsT apa 6 E xov T. jiexpsixw auxov xaxd xov 
H- 6 E apa xov H TtoXXaTtXaaidaag xov T TtETKHirpcev. dXXa 
(i/)v 8id xo Tipo xouxou xal 6 A xov B TioXXaTtXaaidaac; xov 
r TCETioirjxev. 6 apa ex xwv A, B laoc, saxl xco ex x£5v E, H. 
saxiv apa tbc; 6 A Ttpoc; xov E, 6 H Ttpoc; xov B. oi 8e A, E 
TcpGxoi, oi 8s Ttpwxoi xal eXd)(ioxoL, oi 8s eXd)(iaxoi dpi%ol 



measures D the same number of times as B (measures) 
E [Prop. 7.15]. And the unit A measures D according to 
the units in it. Thus, B also measures E according to the 
units in D. Hence, the lesser (number) B measures the 
greater E according to some existing number among the 
proportional numbers (namely, D). 

Corollary 

And (it is) clear that what(ever relative) place the 
measuring (number) has from the unit, the (number) 
according to which it measures has the same (relative) 
place from the measured (number), in (the direction of 
the number) before it. (Which is) the very thing it was 
required to show. 

Proposition 12 

If any multitude whatsoever of numbers is continu- 
ously proportional, (starting) from a unit, then however 
many prime numbers the last (number) is measured by, 
the (number) next to the unit will also be measured by 
the same (prime numbers). 

A i 1 E i 1 

B i 1 F i 1 

C i 1 G' 1 

D 1 Hi 1 

Let any multitude whatsoever of numbers, A, B, C, 
D, be (continuously) proportional, (starting) from a unit. 
I say that however many prime numbers D is measured 
by, A will also be measured by the same (prime numbers). 

For let D be measured by some prime number E. I 
say that E measures A. For (suppose it does) not. E is 
prime, and every prime number is prime to every num- 
ber which it does not measure [Prop. 7.29]. Thus, E 
and A are prime to one another. And since E measures 
D, let it measure it according to F. Thus, E has made 
D (by) multiplying F. Again, since A measures D ac- 
cording to the units in C [Prop. 9.11 corr.], A has thus 
made D (by) multiplying C. But, in fact, E has also 
made D (by) multiplying F. Thus, the (number cre- 
ated) from (multiplying) A, C is equal to the (number 
created) from (multiplying) E, F. Thus, as A is to E, 
(so) F (is) to C [Prop. 7.19]. And A and E (are) prime 
(to one another), and (numbers) prime (to one another 
are) also the least (of those numbers having the same ra- 
tio as them) [Prop. 7.21], and the least (numbers) mea- 
sure those (numbers) having the same ratio as them an 
equal number of times, the leading (measuring) the lead- 



262 



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ELEMENTS BOOK 9 



[isxpouai xou<; xov auxov Xoyov £)(ovxac; auxolc; iadxig 5 xe 
f)you|ievo<; xov fjyou^ievov xal 6 eit6\±evoq xov feraD^isvov 
^expei apa 6 E xov B. jiexpeixco auxov xaxd xov 9- 6 E apa 
xov O TioXXarcXaaidaat; xov B KETioirjxev. dXXd jir]v xal 6 A 
sauxov 7ioXXanXaaidaa<; xov B 7t£7toiT)xev 6 apa ex xwv E, 
9 iao<; eaxl xo duo xou A. eaxiv apa d><; 6 E npoc, xov A, 6 A 
■rcpoc xov 0. oi Se A, E Ttpwxoi, oi 8e TtpGxoi xal eXd)(iaxoi, oi 
8s sXdxiaxoi (iexpouai xouc; xov auxov Xoyov £)(ovxac; iadxig 
6 fjyou^evog xov fjyoujisvov xal 6 sk6\xsmoc, xov eho^isvov 
[iexpsl apa 6 E xov A w<; r\\o\j\ievoz fjyoujievov. dXXd [i/]v 
xal ou (Jtexpel - onsp dBuvaxov. oux apa oi E, A Tipwxoi npbz 
dXXrjXouc; eioiv. auvdexoi apa. oi 8s auvdexoi uno [upcbxou] 
dpii9(io0 xivog ^texpouvxai. xal snsl 6 E upaixoc; unoxeixai, 6 
6e upwxoc; Otto exepou apiS^ou ou [isxpelxai fj ucp' eauxou, 6 
E apa xou? A, E ^isxpeT- uoxe 6 E xov A ^lexpeT. ^lexpei 8s 
xal xov A - 6 E apa xou<; A, A ^.expeT. 6^io[«<; 8f) Ssi^o^iev, 
oxi ucp' oaov dv 6 A upwxov dpid^iGv [isxprjxai, uuo iSv 
auxCSv xal 6 A [isxprydrjaexai- orcep e8si 8ei<;ai. 



ing, and the following the following [Prop. 7.20] . Thus, 
E measures C. Let it measure it according to G. Thus, 
E has made C (by) multiplying G. But, in fact, via the 
(proposition) before this, A has also made C (by) multi- 
plying B [Prop. 9.11 corr.]. Thus, the (number created) 
from (multiplying) A, B is equal to the (number created) 
from (multiplying) E, G. Thus, as A is to E, (so) G 
(is) to B [Prop. 7.19]. And A and E (are) prime (to 
one another), and (numbers) prime (to one another are) 
also the least (of those numbers having the same ratio 
as them) [Prop. 7.21], and the least (numbers) measure 
those (numbers) having the same ratio as them an equal 
number of times, the leading (measuring) the leading, 
and the following the following [Prop. 7.20]. Thus, E 
measures B. Let it measure it according to H. Thus, 
E has made B (by) multiplying H. But, in fact, A has 
also made B (by) multiplying itself [Prop. 9.8]. Thus, 
the (number created) from (multiplying) E, H is equal 
to the (square) on A. Thus, as E is to A, (so) A (is) 
to H [Prop. 7.19]. And A and E are prime (to one an- 
other), and (numbers) prime (to one another are) also 
the least (of those numbers having the same ratio as 
them) [Prop. 7.21], and the least (numbers) measure 
those (numbers) having the same ratio as them an equal 
number of times, the leading (measuring) the leading, 
and the following the following [Prop. 7.20]. Thus, E 
measures A, as the leading (measuring the) leading. But, 
in fact, (E) also does not measure (A). The very thing 
(is) impossible. Thus, E and A are not prime to one 
another. Thus, (they are) composite (to one another). 
And (numbers) composite (to one another) are (both) 
measured by some [prime] number [Def. 7.14]. And 
since E is assumed (to be) prime, and a prime (number) 
is not measured by another number (other) than itself 
[Def. 7.11], E thus measures (both) A and E. Hence, E 
measures A. And it also measures D. Thus, E measures 
(both) A and D. So, similarly, we can show that however 
many prime numbers D is measured by, A will also be 
measured by the same (prime numbers). (Which is) the 
very thing it was required to show. 



'Edv dno [iovd8o<; 6noooiouv dpid^iol tEjfi dvdXoyov 
Goiv, 6 5e \xeto. xr]v ^iovd8a npfiixoc; fj, 6 [leyiaxog un 
ouSevoc; [aXXou] jjiexpiq'drpsxai napsE, x«v UTtap/ovxiov ev 
xou; dvdXoyov dpid^iou;. 

'Eaxwaav and ^lovdSoc; oiioaoiouv apidfjioi e^rj<; dvdXoyov 
oi A, B, T, A, 6 8e ^texd x/]v ^tovd8a 6 A TtpGxoc; eaxw 
Xcyw, oxi 6 ^tsyiaxoc; auxfiv 6 A bn ouSevoc; aXXou [jle- 
xp/]i9r)Gexai Ttapec; x«v A, B, T. 



Proposition 13 

If any multitude whatsoever of numbers is continu- 
ously proportional, (starting) from a unit, and the (num- 
ber) after the unit is prime, then the greatest (number) 
will be measured by no [other] (numbers) except (num- 
bers) existing among the proportional numbers. 

Let any multitude whatsoever of numbers, A, B, C, 
D, be continuously proportional, (starting) from a unit. 
And let the (number) after the unit, A, be prime. I say 



263 



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B 



r h 

Ah 



Hi ' 

©i ' 



El yap 8uvax6v, [isxpeio'dw utio tou E, xal 6 E ^ir)8evl 
xfiv A, B, r eaxw 6 auxoc;. cpavepov 8rj, oxi 6 E Tiptoxoc; 
oux eaxiv. el yap 6 E Tipfixoc; eaxi xal (jiexpel xov A, xal xov 
A [i£Tpr\o£i Tipaixov ovxa [if] uv auxai 6 auxoc;' ouep eaxlv 
d8uvaxov. oux apa 6 E Tipfixoc; eaxiv. auvdexoc; apa. Tide; 
8e auvdexoc; dpidjioc; utio npwxou xivoc; dpidjiou [iexpeTxar 
6 E apa utio Tipwxou xivoc; dpidjiou (icxpelxai. Xeyio 8r|, oxi 
On' ouSevoc; aXXou Tiptoxou [icxprydriaexai tiX/jv xou A. el yap 
Ocp° exepou [Jiexpelxai 6 E, 6 8s E xov A [iexpel, xdxelvoc; apa 
xov A [L£Tpr\oev a>axe xal xov A (iexpr]aei Tipaixov ovxa \ir\ 
&>v auxfi 6 auxoc;' onep eaxlv d8uvaxov. 6 A apa xov E 
(jiexpel. xal inz\ 6 E xov A [iexpel, [lexpeixw auxov xaxd xov 
Z. Xeyco, oxi 6 Z ouSevl tuv A, B, T eaxiv 6 auxoc;. el yap 6 
Z evl xaiv A, B, T eaxiv 6 auxoc; xal [iexpel xov A xaxd xov 
E, xal elg apa xwv A, B, T xov A [iexpel xaxd xov E. dXXa 
sic, xwv A, B, r xov A [iexpel xaxd xiva xwv A, B, T- xal 6 
E apa evl xwv A, B, T eaxiv 6 auxoc;' onep oux UTioxeixai. 
oux apa 6 Z evl xwv A, B, T eaxiv 6 auxoc;. 6[ioia>c; Br| 
8eii;o[i£v, oxi [lexpelxai 6 Z utio xou A, Beixvuvxec; TidXiv, 
oxi 6 Z oux eaxi Tipcoxoc;. ei yap, xal [iexpel xov A, xal xov 
A [iexpr|aei upSSxov ovxa \xf\ d>v auxcS 6 auxoc;' oicep eaxlv 
dSuvaxov oux apa itpfixoc; eaxiv 6 Z' auvdexoc; apa. dnac; 
8e auvdexoc; dpi'djioc; utio Ttpwxou xivoc; dpid[iou (lexpelxai' 6 
Z apa utio upcoxou xivoc; dpi-djiou [lexpelxai. Xeyco 8rj, oxi ucp' 
exepou jcpwxou ou [i£xp/]iL>r]aexai tiX/]v xou A. ei yap exepoc; 
xic; Tipwxoc; xov Z (iexpel, 6 8e Z xov A jiexpel, xdxelvoc; 
apa xov A [iexpf|aei' waxe xal xov A [lexprpei Tipwxov ovxa 
\ir\ <2>v auxfi) 6 auxoc;' oTiep eaxlv dBuvaxov. 6 A apa xov Z 
[iexpel. xal end 6 E xov A [iexpel xaxd xov Z, 6 E apa xov 
Z TtoXXauXaaidaac; xov A TiCTioirjxev. dXXa [irjv xal 6 A xov 
r TioXXaTiXaaidaac; xov A TiCTioirjxev 6 apa ex xwv A, T 
Xaoc, eaxi to ex xCSv E, Z. dvdXoyov apa eaxlv wc; 6 A Tipoc; 
xov E, ouxcoc; 6 Z Tipoc; xov T. 6 8e A xov E [iexpel' xal 6 Z 
apa xov r [iexpel. [lexpeixw auxov xaxd xov H. o^ioiwc; 8r) 
8eic;o[iev, oxi 6 H ou8evl xwv A, B eaxiv 6 auxoc;, xal oxi 
[lexpelxai utio xou A. xal enel 6 Z xov V [iexpel xaxd xov H, 
6 Z apa xov H TioXXauXaaidaac; xov V 7ieiio(r]xev. dXXa \ir\\> 
xal 6 A xov B TioXXanXaaidaac; xov T 7i£7iolr)xev 6 apa ex 
xov A, B laoc; eaxl xw ex xwv Z, H. dvdXoyov apa <!><; 6 A 
Tipoc; xov Z, 6 H Tipoc; xov B. [xexpeT Be 6 A xov Z- (iexpel 
apa xal 6 H xov B. (xexpeixw auxov xaxd xov 0. 6(ioiw<; 8r) 
8eic;o[iev, oxi 6 9 ifi A oux eaxiv 6 auxoc;. xal cticI 6 H xov 



that the greatest of them, D, will be measured by no other 
(numbers) except A, B, C. 

A' 1 E i 1 



F ^ 



For, if possible, let it be measured by E, and let E not 
be the same as one of A, B, C. So it is clear that E is 
not prime. For if E is prime, and measures D, then it will 
also measure A, (despite ^4) being prime (and) not being 
the same as it [Prop. 9.12]. The very thing is impossible. 
Thus, E is not prime. Thus, (it is) composite. And every 
composite number is measured by some prime number 
[Prop. 7.31]. Thus, E is measured by some prime num- 
ber. So I say that it will be measured by no other prime 
number than A. For if E is measured by another (prime 
number), and E measures D, then this (prime number) 
will thus also measure D. Hence, it will also measure A, 
(despite A) being prime (and) not being the same as it 
[Prop. 9.12]. The very thing is impossible. Thus, A mea- 
sures E. And since E measures D, let it measure it ac- 
cording to F. I say that F is not the same as one of A, B, 
C. For if F is the same as one of A, B, C, and measures 
D according to E, then one of A, B, C thus also measures 
D according to E. But one of A, B, C (only) measures 
D according to some (one) of A, B, C [Prop. 9.11]. And 
thus E is the same as one of A, B, C. The very oppo- 
site thing was assumed. Thus, F is not the same as one 
of A, B, C. Similarly, we can show that F is measured 
by A, (by) again showing that F is not prime. For if (F 
is prime), and measures D, then it will also measure A, 
(despite A) being prime (and) not being the same as it 
[Prop. 9.12]. The very thing is impossible. Thus, F is 
not prime. Thus, (it is) composite. And every composite 
number is measured by some prime number [Prop. 7.31]. 
Thus, F is measured by some prime number. So I say 
that it will be measured by no other prime number than 
A. For if some other prime (number) measures F, and 
F measures D, then this (prime number) will thus also 
measure D. Hence, it will also measure A, (despite A) 
being prime (and) not being the same as it [Prop. 9.12]. 
The very thing is impossible. Thus, A measures F. And 
since E measures D according to F, E has thus made 
D (by) multiplying F. But, in fact, A has also made D 
(by) multiplying C [Prop. 9.11 corr.]. Thus, the (number 
created) from (multiplying) A, C is equal to the (number 
created) from (multiplying) E, F. Thus, proportionally, 
as A is to E, so F (is) to C [Prop. 7.19]. And A measures 



264 



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ELEMENTS BOOK 9 



B jjiexpsT xaxd xov 0, 6 H dpa xov TtoXXaTiXaa(.daa<; xov 
B 7i£iio(r]xev. dXXd ^f)V xal 6 A eauxov TioXXaTcXaaidaac; 
xov B TiETtoirjxev 6 dpa utio 6, H iaoz eaxl iu dTio xou A 
Teipaywvw' eaxiv dpa u<;6 8 Tipoc; xov A, 6 A Tip6<; tov H. 
[icxpeT 8e 6 A xov H- [icxpeT dpa xal 6 O xov A npoxov ovxa 
\ir\ &v auxSS 6 auxoc oTiep dxoTiov. oux dpa 6 ^eyiaxoc; 6 
A utio exepou dpidjiou ueTpryft^asxai Tiape^ x«v A, B, T- 
ojiep sSei 5sT5ai. 



E. Thus, F also measures C. Let it measure it according 
to G. So, similarly, we can show that G is not the same 
as one of A, B, and that it is measured by A. And since 
F measures C according to G, F has thus made C (by) 
multiplying G. But, in fact, A has also made C (by) mul- 
tiplying B [Prop. 9.11 corr.]. Thus, the (number created) 
from (multiplying) A, B is equal to the (number created) 
from (multiplying) F, G. Thus, proportionally, as A (is) 
to F, so G (is) to B [Prop. 7.19]. And A measures F. 
Thus, G also measures B. Let it measure it according to 
H. So, similarly, we can show that H is not the same as 
A. And since G measures B according to H, G has thus 
made B (by) multiplying H. But, in fact, A has also made 
B (by) multiplying itself [Prop. 9.8]. Thus, the (number 
created) from (multiplying) H, G is equal to the square 
on A. Thus, as H is to A, (so) A (is) to G [Prop. 7.19]. 
And A measures G. Thus, H also measures A, (despite 
A) being prime (and) not being the same as it. The very 
thing (is) absurd. Thus, the greatest (number) D cannot 
be measured by another (number) except (one of) A, B, 
C. (Which is) the very thing it was required to show. 



18'. 

'Edv skctyioToz dpidjioe; utio icpwxwv dpid^wv ^.expfjxai, 
bn ou8evo<; dXXou icpwxou dpid^ou jiexp/jOiqaexai Tiape?; 
xwv 15 dpxfjt; ^lexpouvxtov. 



B ^ 

r h 



Ah 



2' — i 

'EXd)(ioxoc; yap dpn&jioc; 6 A utio icpoxcov dpidjifiv xwv 
B, T, A ^expeicrdw Xeyw, oxi 6 A Ok' ouSevcx; dXXou 
itpwxou dpiiSjioO (lexprydrpexai uape^ xaiv B, T, A. 

EE yap Suvaxov, (lexpeio-dio utio Tipcoxou xou E, xal 6 
E [i.r)8evl xSv B, T, A eaxco 6 auxog. xal fenel 6 E xov 
A [iexpeT, (icxpeixio auxov xaxa xov Z' 6 E dpa xov Z 
TtoXXaTiXaaidaag xov A icsitoi/jxev. xal jiexpelxai 6 A utio 
Ttpcjxcov dpii9[i£iv xcov B, T, A. edv Be Buo dpid^iol tioX- 
XaTcXaoidaavxs<; dXXf]Xou<; koiGSoi xiva, xov Be yevo^ievov 
ec; auxwv fiexpfj xlc Ttptoxoc api'&y.oz, xal eva xwv fE, dp^fjc; 
\iSTprpei' o\ B, T, A dpa eva xwv E, Z [lexprpouaiv. xov 
^xev ouv E ou ^lexpiqaoumv 6 yap E Tcpox6<; eaxi xal ouBevl 
xwv B, T, A 6 auxo<;. xov Z dpa ^texpouaiv eXdaaova ovxa 
xou A' oTtep dBuvaxov. 6 yap A UTioxeixai eXd)(io"xoc utio 
xwv B, T, A ^.expou^tevot;. oux dpa xov A \iSTpr\ae\. TipGxoc; 
dpid^icx; Tiape^ xwv B, T, A- OTiep eSei BeT^ai. 



Proposition 14 

If a least number is measured by (some) prime num- 
bers then it will not be measured by any other prime 
number except (one of) the original measuring (num- 
bers). 

A 1 B i 1 



F h 



Dh 



H 



For let A be the least number measured by the prime 
numbers B, C, D. I say that A will not be measured by 
any other prime number except (one of) B, C, D. 

For, if possible, let it be measured by the prime (num- 
ber) E. And let E not be the same as one of B, C, D. 
And since E measures A, let it measure it according to F. 
Thus, E has made A (by) multiplying F. And A is mea- 
sured by the prime numbers B, C, D. And if two num- 
bers make some (number by) multiplying one another, 
and some prime number measures the number created 
from them, then (the prime number) will also measure 
one of the original (numbers) [Prop. 7.30]. Thus, B, C, 
D will measure one of E, F. In fact, they do not measure 
E. For E is prime, and not the same as one of B, C, D. 
Thus, they (all) measure F, which is less than A. The 
very thing (is) impossible. For A was assumed (to be) the 
least (number) measured by B, C, D. Thus, no prime 



265 



ETOIXEIfiN fl'. 



ELEMENTS BOOK 9 



is'. 

'Edv xpeTc; dpid^toi ec;rj<; dvaXoyov Saiv eXd)(iaxoi xwv 
xov auxov Xoyov e)(6vxwv auxolc;, Buo ottoioiouv auv- 
xcdevxec; Tcpoc xov Xomov upwxoi Eiaiv. 

A E 2 

A i 1 i 1 1 

B i 1 

r i 1 

'Eaxwaav xpeu; apid^oi i^f\z dvaXoyov eXd)(iaxoi xSv 
xov auxov Xoyov e)(6vxwv auxolc; oi A, B, E Xeyw, oxi xwv 
A, B, T 5uo otioioioDv auvxcdevxeg npbc, xov Xomov upwxoi 
eiaiv, oi [lev A, B Tipot; xov T, oi 8s B, T npbc, xov A xai 
exi oi A, r 7tp6<; xov B. 

EiXr ! )(p , dwaav yap eXd/iaxoi apidjjioi xwv xov auxov 
Xoyov e^ovxwv xo ^ -A- 1 B, r Buo oi AE, EZ. cpavepov 
8r), oxi 6 (isv AE eauxov TioXXajcXaaidaac; xov A TteTcoirjxev, 
xov Be EZ TioXXaicXaaidaac; xov B iieTcoir)xev, xai ixi 6 EZ 
eauxov TtoXXajcXaaidaac; xov T KSTioirjXEv. xai eicei oi AE, 
EZ eXdxiaxoL eiaiv, itpwxoi npoc; dXXr|Xou<; eiaiv. eav Be 
Buo dpi%oi itpwxoi upog dXXrjXouc; waiv, xai auva|icp6xepoc; 
npoc exdxepov Ttpwxoc; eaxiv xai 6 AZ dpa upoc; exdxepov 
xwv AE, EZ npwxoc; eaxiv. dXXd xai 6 AE npbc, xov 
EZ itpwxoc; eaxiv oi AZ, AE dpa npoc; xov EZ Ttpwxoi eiaiv. 
eav Be Buo dpidjioi upog xiva dpL^jiov itpwxoi waiv, xai 6 
zl, auxwv yevo^xevoc; 7cp6c; xov Xomov itpwxoc; eaxiv waxe 
6 ex xwv ZA, AE npoc; xov EZ Ttpwxoc; eaxiv waxe xai 6 
ex xwv ZA, AE icpot; xov arco xou EZ Tcpwxoc; eaxiv. [eav 
yap Buo dpii9(ioi Ttpwxoi Ttpoc; dXXr]Xouc; waiv, 6 ex xou tvbc, 
auxwv yevo^ievoc; Ttpoc; xov Xoitiov Ttpwxoc; eaxiv]. dXX' 6 
ex xwv ZA, AE 6 dno xou AE eaxi jiexa xou ex xwv AE, 
EZ- 6 dpa duo xou AE jiexa xou ex xwv AE, EZ Ttpoc; xov 
aTio xou EZ Ttpwxoc; eaxiv. xai eaxiv 6 uev duo xou AE 
6 A, 6 Be ex xwv AE, EZ 6 B, 6 Be duo xou EZ 6 E oi 
A, B dpa auvxei!)evxe<; Ttpoc; xov T Ttpwxoi eiaiv. ojioiwc; Srj 
8eic;ouev, oxi xai oi B, T npbc, xov A Ttpwxoi eiaiv. Xeyw 
Srj, oxi xai oi A, T npbc, xov B Ttpwxoi eiaiv. CTtei yap 6 AZ 
Ttpoc; exdxepov xwv AE, EZ Ttpwxoc eaxiv, xai 6 aTto xou 
AZ npbc, xov ex xwv AE, EZ Ttpwxoc eaxiv. dXXd xw drco 
xou AZ i'aoi eiaiv oi aTto xwv AE, EZ jiexa xou Bic ex xwv 
AE, EZ- xai oi aTto xwv AE, EZ dpa (icxa xou Si? bnb xwv 
AE, EZ Ttpoc xov (mo xwv AE, EZ Ttpwxoi [eiai]. BieXovxi 
oi duo xwv AE, EZ ^icxa xou aTtac; bnb AE, EZ Ttpoc xov 
Oito AE, EZ Ttpwxoi eiaiv. exi BieXovxi oi aTto xwv AE, EZ 
dpa Ttpoc xov utio AE, EZ Ttpwxoi eiaiv. xai eaxiv 6 jxev 



number can measure A except (one of) B, C, D. (Which 
is) the very thing it was required to show. 

Proposition 15 

If three continuously proportional numbers are the 
least of those (numbers) having the same ratio as them 
then two (of them) added together in any way are prime 
to the remaining (one) . 

D E F 

A i 1 i 1 1 

Bi ' 

C' ' 

Let A, B, C be three continuously proportional num- 
bers (which are) the least of those (numbers) having the 
same ratio as them. I say that two of A, B, C added to- 
gether in any way are prime to the remaining (one), (that 
is) A and B (prime) to C, B and C to A, and, further, A 
and C to B. 

Let the two least numbers, DE and EF, having the 
same ratio as A, B, C, have been taken [Prop. 8.2]. 
So it is clear that DE has made A (by) multiplying it- 
self, and has made B (by) multiplying EF, and, fur- 
ther, EF has made C (by) multiplying itself [Prop. 8.2]. 
And since DE, EF are the least (of those numbers hav- 
ing the same ratio as them), they are prime to one an- 
other [Prop. 7.22]. And if two numbers are prime to 
one another then the sum (of them) is also prime to each 
[Prop. 7.28]. Thus, DF is also prime to each of DE, EF. 
But, in fact, DE is also prime to EF. Thus, DF, DE 
are (both) prime to EF. And if two numbers are (both) 
prime to some number then the (number) created from 
(multiplying) them is also prime to the remaining (num- 
ber) [Prop. 7.24]. Hence, the (number created) from 
(multiplying) FD, DE is prime to EF. Hence, the (num- 
ber created) from (multiplying) FD, DE is also prime 
to the (square) on EF [Prop. 7.25]. [For if two num- 
bers are prime to one another then the (number) created 
from (squaring) one of them is prime to the remaining 
(number).] But the (number created) from (multiplying) 
FD, DE is the (square) on DE plus the (number cre- 
ated) from (multiplying) DE, EF [Prop. 2.3]. Thus, the 
(square) on DE plus the (number created) from (multi- 
plying) DE, EF is prime to the (square) on EF. And 
the (square) on DE is A, and the (number created) from 
(multiplying) DE, EF (is) B, and the (square) on EF 
(is) C. Thus, A, B summed is prime to C. So, similarly, 
we can show that B, C (summed) is also prime to A. So 
I say that A, C (summed) is also prime to B. For since 



266 



ETOIXEIfiN fl'. 



ELEMENTS BOOK 9 



duo xoO AE 6 A, 6 Be: utio xwv AE, EZ 6 B, 6 8s dtKo xou DF is prime to each of DE, EF then the (square) on DF 
EZ 6 T. ol A, T dpa auvxe'devxec; Tipoc xov B Ttpfixoi elcnv is also prime to the (number created) from (multiplying) 
oTisp eBsi hzXiai. DE, EF [Prop. 7.25]. But, the (sum of the squares) on 

DE, EF plus twice the (number created) from (multiply- 
ing) DE, EF is equal to the (square) on DF [Prop. 2.4]. 
And thus the (sum of the squares) on DE, EF plus twice 
the (rectangle contained) by DE, EF [is] prime to the 
(rectangle contained) by DE, EF. By separation, the 
(sum of the squares) on DE, EF plus once the (rect- 
angle contained) by DE, EF is prime to the (rectangle 
contained) by DE, EFJ Again, by separation, the (sum 
of the squares) on DE, EF is prime to the (rectangle 
contained) by DE, EF. And the (square) on DE is A, 
and the (rectangle contained) by DE, EF (is) B, and 
the (square) on EF (is) C. Thus, A, C summed is prime 
to B. (Which is) the very thing it was required to show. 

t Since if a f3 measures a 2 + (3 2 + 2 a (3 then it also measures a 2 + ffi + a (3, and vice versa. 



If'. 

'Edv Buo dpid|jiol Ttpwxoi. 7ip6<; dXXr|Xouc Soiv, oux eaxai 
cbc; 6 npcbxoc; Ttpoc; xov Bsuxspov, ouxtoc; 6 Beuxepoe; 7ip6<; 
aXXov xivd. 

A i ' 

B i ' 

r i 1 

Auo yap dpu^oi oi A, B nptoxoi npoc; dXXr|Xouc; eoxto- 
aocv Xeyw, oxi oux eaxiv cbc; 6 A npoc, xov B, ouxw<; 6 B 
7ip6<; aXXov xivd. 

El yap Buvaxov, eaxco cbc; 6 A 7ip6<; xov B, 6 B Ttpoe; 
xov r. ol Be A, B upOxoi, oi Be Ttpcbxoi xal eXd)(iaxoi, oi Be 
sXa/iaxoi dpidjiol ^.expoOai xouc; xov auxov Xoyov exovxag 
iadxic; 8 xe f}yo'j\±evoz xov f]Y ou ^ £vov xal ° etco^isvoc, xov 
£ttojj£vov (jiexpsl dpa 6 A xov B cbc, r)you[jievoc; fiyoujievov. 
(iexpeT Be xal eauxov 6 A dpa xou<; A, B ^lexpei itpcbxouc; 
ovxac; 7ip6<; dXXfjXouc ouep dxorcov. oux apa saxai cbc; 6 A 
npoc, xov B, ouxtoc; 6 B Ttpoc; xov T- 6mp eSei Belc^ai. 



'Edv Soiv oaoiBrjTioxouv dpid^toi e^rjc; dvdXoyov, oi Be 
axpoi auxcbv upwxoi rcpoc; dXXrjXouc; Saiv, oux eaxai cbc; 6 
upCSxoc; Tipoc; xov Beruxepov, ouxck 6 eo)(axo<; Ttpoc; dXXov 



Proposition 16 

If two numbers are prime to one another then as the 
first is to the second, so the second (will) not (be) to some 
other (number) . 

A' 1 

B 1 

C' ' 

For let the two numbers A and B be prime to one 
another. I say that as A is to B, so B is not to some other 
(number). 

For, if possible, let it be that as A (is) to B, (so) 
B (is) to C. And A and B (are) prime (to one an- 
other). And (numbers) prime (to one another are) also 
the least (of those numbers having the same ratio as 
them) [Prop. 7.21]. And the least numbers measure 
those (numbers) having the same ratio (as them) an 
equal number of times, the leading (measuring) the lead- 
ing, and the following the following [Prop. 7.20] . Thus, 
A measures B, as the leading (measuring) the leading. 
And (A) also measures itself. Thus, A measures A and B, 
which are prime to one another. The very thing (is) ab- 
surd. Thus, as A (is) to B, so B cannot be to C. (Which 
is) the very thing it was required to show. 

Proposition 17 

If any multitude whatsoever of numbers is continu- 
ously proportional, and the outermost of them are prime 
to one another, then as the first (is) to the second, so the 



267 



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ELEMENTS BOOK 9 



xivd. 

'Eaxwaav 6aoi8r]7ioTouv api'd^oi e^fj<; dvdXoyov oi A, 
B, T, A, ol 8e dxpoi auxfiv oi A, A npoxoi npoc dXXr|Xouc; 
eaxcoaav Xeyco, §xi oux eaxiv cb<; 6 A upoc; xov B, ouxok 6 
A rcpoc; dXXov xivd. 

A' 1 

B ' ' 

r ' 1 

A i ' 

Ei ' 

El yap 8uvax6v, eaxco foe. 6 A npoz xov B, ouxox 6 A 
7ip6<; xov E- evaXXd?; dpa eaxiv 6? 6 A Ttpoc xov A, 6 B Ttpoc; 
xov E. oi 8e A, A itpcoxoi, oi 8e TtpGxoi xai tXayiaxoi, oi 8s 
eXd)(iaxoi apidjiol ^texpouai xou<; xov auxov Xoyov e)(ovxa<; 
iadxi<; o xe f]you^ievo<; TOV /]You^evov xa ' 1 ° i^o[ievoc, xov 
CTtojievov. [jiexpei dpa 6 A xov B. xai eaxiv foe, 6 A npbe 
xov B, 6 B icpoc^ xov T. xai 6 B dpa xov T jxexpeT - &axe 
xai 6 A xov T jiexpei. xai inei eaxiv (be; 6 B npog xov T, 
6 r Kpoc; xov A, ^.expeT Be 6 B xov T, ^.expeT dpa xai 6 T 
xov A. dXX' 6 A xov T E^expei' waxe 6 A xai xov A ^expert. 
^expeT 8s xai eauxov. 6 A dpa xou<; A, A ^expeT n:pa>xou<; 
ovxa<; 7ip6<; dXXf|Xou<; - oiiep eaxiv dBuvaxov. oux dpa eaxai 
foe, 6 A 7ip6<; xov B, oux«<; 6 A npbe dXXov xivd- ouep e8ei 
8eT?ai. 



IT]'. 

Auo dpi%£5v Bo'devxwv emaxetjjaa'dai, el Buvaxov eaxiv 
auxolg xpixov dvdXoyov rcpoaeupeTv. 

a i 1 r i 1 

B i 1 A i 1 

"Eaxwaav oi BoiJevxet; 8uo dprdjioi oi A, B, xai 
8eov eaxw eraaxe^aa'dai, si 8uvax6v eaxiv auxolc; xpixov 
dvdXoyov Ttpoaeupefv. 

Oi 8r) A, B ffzoi Tipfixoi upog dXXr]Xouc; eiaiv fj ou. xai si 
Tipwxoi npbe. dXXf]Xouc; eiaiv, SeBeixxai, oxi dBuvaxov eaxiv 
auxou; xpixov dvdXoyov TtpoaeupeTv. 

AXXd 8r) \±r\ eaxtoaav oi A, B rcpfiixoi npbe dXXrjXouc;, 
xai 6 B eauxov TioXXanXaaidaag xov T noieixw. 6 A §r| xov 
T f]xoi jiexpei f] ou jiexpei. |iexpeix« npoxepov xaxd xov A- 
6 A dpa xov A TroXXanXaaidaac; xov T nenoi/jxev. dXXa ^ir)v 
xai 6 B eauxov noXXaTiXaaidaag xov T nz%oir\xev b dpa 



last will not be to some other (number). 

Let A, B, C, D be any multitude whatsoever of con- 
tinuously proportional numbers. And let the outermost 
of them, A and D, be prime to one another. I say that as 
A is to B, so D (is) not to some other (number) . 

A' 1 

B ' 1 

Ci ' 

D' ' 

Ei ' 

For, if possible, let it be that as A (is) to B, so D 
(is) to E. Thus, alternately, as A is to D, (so) B (is) 
to E [Prop. 7.13]. And A and D are prime (to one 
another). And (numbers) prime (to one another are) 
also the least (of those numbers having the same ra- 
tio as them) [Prop. 7.21]. And the least numbers mea- 
sure those (numbers) having the same ratio (as them) an 
equal number of times, the leading (measuring) the lead- 
ing, and the following the following [Prop. 7.20] . Thus, 
A measures B. And as A is to B, (so) B (is) to C. Thus, B 
also measures C. And hence A measures C [Def. 7.20]. 
And since as B is to C, (so) C (is) to D, and B mea- 
sures C, C thus also measures D [Def. 7.20] . But, A was 
(found to be) measuring C. And hence A also measures 
D. And (A) also measures itself. Thus, A measures A 
and D, which are prime to one another. The very thing is 
impossible. Thus, as A (is) to B, so D cannot be to some 
other (number) . (Which is) the very thing it was required 
to show. 

Proposition 18 

For two given numbers, to investigate whether it is 
possible to find a third (number) proportional to them. 

A i 1 C' ' 

B i 1 Di 1 

Let A and B be the two given numbers. And let it 
be required to investigate whether it is possible to find a 
third (number) proportional to them. 

So A and B are either prime to one another, or not. 
And if they are prime to one another then it has (already) 
been show that it is impossible to find a third (number) 
proportional to them [Prop. 9.16]. 

And so let A and B not be prime to one another. And 
let B make C (by) multiplying itself. So A either mea- 
sures, or does not measure, C. Let it first of all measure 
(C) according to D. Thus, A has made C (by) multiply- 



268 



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ELEMENTS BOOK 9 



ex xfiv A, A iaoc; eaxi xfi dmo xou B. eaxiv apa &>z 6 A 
7ip6<; xov B, 6 B Tipoc; xov A- xdig A, B apa xpixoc; dpid^toc; 
dvdXoyov Tipoar]upr)xai 6 A. 

AXXa 8f] [i/j [icxpeixto 6 A xov E Xeyw, oxi xolc; A, B 
dSuvaxov eaxi xpixov dvdXoyov TtpoaeupeTv dpidjiov. et yap 
Buvaxov, n;poar]upr]a , d« 6 A. 6 apa ex xwv A, A Iaoc eaxi 
x£> duo xou B. 6 8e and xou B eaxiv 6 T' 6 apa ex xwv A, 
A iaoc eaxi xC) T. wote 6 A xov A TtoXXaitXaaidaac xov 
r 7te7tcu7]xev 6 A apa xov T ^texpeT xaxa xov A. dXXa (i/]v 
UTtoxeixai xal {jltj ^.expwv orcep axoitov. oux apa Buvaxov 
eaxi xoi"c A, B xpixov dvdXoyov Ttpoaeupelv dpid^ov, oxav 
6 A xov T [jlt) nexpfj- oitep e8ei 8el5ai. 



10'. 

Tpiwv dpnD^iwv BoiSevxMv eTiiaxe<j;aaiL>ai, iraxe Buvaxov 
eaxiv auxoic xexapxov dvdXoyov Ttpoaeupeiv. 

A' 1 

B ' ' 

r ' ' 

A i 1 

E' ' 

"Eaxwaav oi So-devxec xpeTc dpi'd^oi oi A, B, T, xal Seov 
eaxto £TuaxecJ>acrdai, Ttoxe Buvaxov eaxiv auxoic xexapxov 
dvdXoyov Ttpoaeupeiv. 

"Hxoi ouv oux eiaiv ec^fjc dvdXoyov, xal oi axpoi auxfiv 
Tipwxoi Tipoc dXXf|Xouc eiaiv, fj ec;fjc eiaiv dvdXoyov, xal oi 
axpoi auxwv oux eiai Kpwxoi Tipoc dXXr|Xouc, fj ouxe e<;rjc 
eiaiv dvdXoyov, ouxe oi axpoi auxwv Tipoxoi Tipoc dXXr|Xouc 
eiaiv, fj xal e^fjc eiaiv dvdXoyov, xal oi axpoi auxfiv TtpGxoi 
Tipoc dXXf]Xouc eiaiv. 

Ei (jiev ouv oi A, B, T ec;rjc eiaiv dvdXoyov, xal oi 
axpoi auxwv oi A, T upGxoi Tipoc dXXf|Xouc eiaiv, SeBeixxai, 
oxi dBuvaxov eaxiv auxoic xexapxov dvdXoyov Tipoaeupeiv 
dpid^iov. [XT] eaxoaav 8/] oi A, B, T ei;rjc dvdXoyov xwv 
dxpwv TtdXiv ovxwv Ttpwxtov Tipoc dXXf|Xouc. Xey«, oxi 
xal ouxwc dSuvaxov eaxiv auxoic xexapxov dvdXoyov Tipo- 
aeupeTv. ei yap 8uvax6v, Tipoaeuprp'dw 6 A, waxe elvai cbc 
xov A Tipoc xov B, xov r Tipoc xov A, xal yeyovexo 6? 6 B 
Tipoc xov r, 6 A Tipoc xov E. xal eitei eaxiv cbc [lev 6 A Tipoc 
xov B, 6 r Tipoc xov A, «c Be 6 B Tipoc xov T, 6 A Tipoc 
xov E, 8i' i'aou apa cbc 6 A Tipoc xov T, 6 T Tipoc xov E. oi 
Be A, T Ttpwxoi, oi Be TipGxoi xal eXd)(iaxoi, oi Be eXd)(iaxoi 



ing D. But, in fact, B has also made C (by) multiplying 
itself. Thus, the (number created) from (multiplying) A, 
D is equal to the (square) on B. Thus, as A is to B, (so) 
B (is) to D [Prop. 7.19]. Thus, a third number has been 
found proportional to A, B, (namely) D. 

And so let A not measure C. I say that it is impossi- 
ble to find a third number proportional to A, B. For, if 
possible, let it have been found, (and let it be) D. Thus, 
the (number created) from (multiplying) A, D is equal to 
the (square) on B [Prop. 7.19]. And the (square) on B 
is C. Thus, the (number created) from (multiplying) A, 
D is equal to C. Hence, A has made C (by) multiplying 
D. Thus, A measures C according to D. But (A) was, in 
fact, also assumed (to be) not measuring (C). The very 
thing (is) absurd. Thus, it is not possible to find a third 
number proportional to A, B when A does not measure 
C. (Which is) the very thing it was required to show. 

Proposition 19 f 

For three given numbers, to investigate when it is pos- 
sible to find a fourth (number) proportional to them. 

A' 1 

B' 1 

C' ' 

D' 1 

Ei 1 

Let A, B, C be the three given numbers. And let it be 
required to investigate when it is possible to find a fourth 
(number) proportional to them. 

In fact, (A, B, C) are either not continuously pro- 
portional and the outermost of them are prime to one 
another, or are continuously proportional and the outer- 
most of them are not prime to one another, or are neither 
continuously proportional nor are the outermost of them 
prime to one another, or are continuously proportional 
and the outermost of them are prime to one another. 

In fact, if A, B, C are continuously proportional, and 
the outermost of them, A and C, are prime to one an- 
other, (then) it has (already) been shown that it is im- 
possible to find a fourth number proportional to them 
[Prop. 9.17]. So let A, B, C not be continuously propor- 
tional, (with) the outermost of them again being prime to 
one another. I say that, in this case, it is also impossible 
to find a fourth (number) proportional to them. For, if 
possible, let it have been found, (and let it be) D. Hence, 
it will be that as A (is) to B, (so) C (is) to D. And let it be 
contrived that as B (is) to C, (so) D (is) to E. And since 



269 



ETOIXEIfiN fl'. 



ELEMENTS BOOK 9 



[iexpouai xou<; tov auxov Xoyov e)(ovxac; 6 xe fjyoujievoc; 
xov rjyoujievov xal 6 CTio^evoc; xov euojievov. jiexpeT apa 6 
A tov T tdc, f]you(jievoc; fjyou^evov. jiexpel 8e xai eauxov 
6 A apa xou<; A, T ^.expeT 7tp«xou<; ovxac; 7tp6<; dXXfiXouc 
oitep eaxiv dSuvaxov. oux apa xou; A, B, T Buvaxov eaxi 
xexapxov dvdXoyov TtpoaeupeTv. 

AXXd §r] TidXiv eaxwaav oi A, B, T e^fj<; dvdXoyov, oi 8s 
A, r (J.r) eaxoaav npwxoi 7tp6<; dXXf|Xou<;. Xeyw, oxi 8uvax6v 
eaxiv auxolc; xexapxov dvdXoyov KpoaeupeTv. 6 yap B xov T 
7toXXaTtXaaidaa<; xov A Ttoieixw 6 A apa xov A fjxoi [icxpel 
rj ou (jiexpel. (lexpeixM auxov Ttpoxepov xaxd xov E- 6 A apa 
xov E TioXXarcXaaidaai; xov A TteTtoirjxev. dXXd [i/]v xal 6 
B xov r TioXXaTiXaaidaac; xov A 7i£Tio(r]xev 6 apa ex xwv 
A, E Iooq eaxi xw ex xwv B, T. dvdXoyov apa [eaxiv] «<; 6 
A 7ipo<; xov B, 6 T 7tp6<; xov E- xou; A, B, T apa xexapxoc; 
dvdXoyov Ttpoa/]upr)xai 6 E. 

AXXd 8r) \ir\ (jiexpeixo 6 A xov A- Xeyw, oxi dSuvaxov 
eaxi xoT<; A, B, T xexapxov dvdXoyov upoaeupelv dpiiJ^ov. 
ei yap 8uvax6v, TipooeuprjO'dcL) 6 E - 6 apa ex xwv A, E Xaoc, 
eaxl xo ex xwv B, T. dXXd 6 ex xwv B, T eaxiv 6 A' xal 
6 ex xfiiv A, E apa iaoc, eaxl to A. 6 A apa xov E uoXXa- 
TiXaaidaac; xov A 7t£TioiT]xev' 6 A apa xov A jiexpeT xaxd xov 
E' waxe ^expeT 6 A xov A. dXXd xal ou jiexpel" ouep axoKov. 
oux apa 8uvdxov eaxi xou; A, B, T xexapxov dvdXoyov n:po- 
aeupeTv dpi , d[i6v, oxav 6 A xov A [irj (iexpfj. dXXd 8r] oi A, B, 
T \±r\T£ kL,ff, eaxwaav dvdXoyov \xr\Te oi axpoi upOxoi npbz 
dXXf]Xou<;. xal 6 B xov T noXXaTiXaaidaa^ xov A uoielxw. 
b\Loitdz 8r] 8ei)(i5f]aexai, oxi ei ^ev (jiexpel 6 A xov A, 8u- 
vaxov eaxiv auxolc; dvdXoyov npoaeupeTv, ei 8e ou jiexpel, 
d8uvaxov ouep e8ei SeT^ai. 



as A is to B, (so) C (is) to D, and as B (is) to C, (so) D 
(is) to E, thus, via equality, as A (is) to C, (so) C (is) to 
[Prop. 7.14]. And A and C (are) prime (to one another). 
And (numbers) prime (to one another are) also the least 
(numbers having the same ratio as them) [Prop. 7.21]. 
And the least (numbers) measure those numbers having 
the same ratio as them (the same number of times), the 
leading (measuring) the leading, and the following the 
following [Prop. 7.20]. Thus, A measures C, (as) the 
leading (measuring) the leading. And it also measures 
itself. Thus, A measures A and C, which are prime to 
one another. The very thing is impossible. Thus, it is not 
possible to find a fourth (number) proportional to A, B, 
C. 

And so let A, B, C again be continuously propor- 
tional, and let A and C not be prime to one another. I 
say that it is possible to find a fourth (number) propor- 
tional to them. For let B make D (by) multiplying C. 
Thus, A either measures or does not measure D. Let it, 
first of all, measure (D) according to E. Thus, A has 
made D (by) multiplying E. But, in fact, B has also made 
D (by) multiplying C. Thus, the (number created) from 
(multiplying) A, E is equal to the (number created) from 
(multiplying) B, C. Thus, proportionally, as A [is] to B, 
(so) C (is) to E [Prop. 7.19]. Thus, a fourth (number) 
proportional to A, B, C has been found, (namely) E. 

And so let A not measure D. I say that it is impossible 
to find a fourth number proportional to A, B, C. For, if 
possible, let it have been found, (and let it be) E. Thus, 
the (number created) from (multiplying) A, E is equal to 
the (number created) from (multiplying) B, C. But, the 
(number created) from (multiplying) B, C is D. And thus 
the (number created) from (multiplying) A, E is equal to 
D. Thus, A has made D (by) multiplying E. Thus, A 
measures D according to E. Hence, A measures D. But, 
it also does not measure (D). The very thing (is) absurd. 
Thus, it is not possible to find a fourth number propor- 
tional to A, B, C when A does not measure D. And so 
(let) A, B, C (be) neither continuously proportional, nor 
(let) the outermost of them (be) prime to one another. 
And let B make D (by) multiplying C. So, similarly, it 
can be show that if A measures D then it is possible to 
find a fourth (number) proportional to (A, B, C), and 
impossible if (A) does not measure (£>). (Which is) the 
very thing it was required to show. 



t The proof of this proposition is incorrect. There are, in fact, only two cases. Either A, B, C are continuously proportional, with A and C prime 
to one another, or not. In the first case, it is impossible to find a fourth proportional number. In the second case, it is possible to find a fourth 
proportional number provided that A measures B times C. Of the four cases considered by Euclid, the proof given in the second case is incorrect, 
since it only demonstrates that if A : B :: C : D then a number E cannot be found such that B : C :: D : E. The proofs given in the other three 



270 



ETOIXEIfiN f)'. 



ELEMENTS BOOK 9 



cases are correct. 



X . 

Oi Tipwxoi dpidjjioi nXeiouc, eiai Travxog tou rcpoxcdevxoc; 
TtXr^ouc; TtptoTGJv dpidjjicov. 

A i 1 Hi 1 

B i 1 

r i 1 

E A 2 

i 1 — i 

"Eaxwaav oi TtpoxeiDevxef; Tipwxoi dpi/d^ol oi A, B, E 
Xeyco, oxi x65v A, B, T TtXeiouc; eiai upaixoi dpi'djioi. 

EiXyjcp'dw yap 6 utto xcov A, B, T eXd)(iaxoc; ^texpoujievog 
xdi eaxco AE, xal upoaxeia^Gj xcp AE ^ovag f) AZ. 6 5r| EZ 
fjxoi Tipfixog eaxiv fj ou. eaxw Ttpoxepov Ttpcoxoc;- eupr][J.evoi 
dpa eial rcpfiixoi dpnJ^iol oi A, B, T, EZ TiXeiouc; xfiiv A, B, 

r. 

AXXd 8/) jirj eaxw 6 EZ upaixog' utto Ttpcjxou dpa xivoc 
dpiiL>|jioO [icxpelxai. [lexpeiadio utio Tipwxou xou H- Xeyw, 
6xi 6 H ouSevl xGv A, B, T eaxiv 6 auxoc;. ei yap Suvaxov, 
eaxco. oi Be A, B, T xov AE jiexpouaiv xal 6 H dpa xov 
AE piexprjoei. ^texpeT Se xal xov EZ- xal Xomf]v xr|v AZ 
^.ovdBa jiexprjoei 6 H dpid^ioc; cov ojtep axoKov. oux dpa 6 
H evi xGv A, B, T eaxiv 6 auxo<;. xal UTtoxeixai 7tp«xo<;. 
eupr)[ievoi dpa eiai upwxoi dpid^ol nkeiouz xoO Ttpoxe , devxo<; 
TtXTydou? xGv A, B, T oi A, B, T, H- onep eBei SeT^ai. 



Proposition 20 

The (set of all) prime numbers is more numerous than 
any assigned multitude of prime numbers. 

A i 1 G' 1 

B 1 

C' 1 

E D F 

i 1 1 

Let A, B, C be the assigned prime numbers. I say that 
the (set of all) primes numbers is more numerous than A, 
B, C. 

For let the least number measured by A, B, C have 
been taken, and let it be DE [Prop. 7.36]. And let the 
unit DF have been added to DE. So EF is either prime, 
or not. Let it, first of all, be prime. Thus, the (set of) 
prime numbers A, B, C, EF, (which is) more numerous 
than A, B, C, has been found. 

And so let EF not be prime. Thus, it is measured by 
some prime number [Prop. 7.31]. Let it be measured by 
the prime (number) G. I say that G is not the same as 
any of A, B, C. For, if possible, let it be (the same). And 
A, B, C (all) measure DE. Thus, G will also measure 
DE. And it also measures EF. (So) G will also mea- 
sure the remainder, unit DF, (despite) being a number 
[Prop. 7.28]. The very thing (is) absurd. Thus, G is not 
the same as one of A, B, C. And it was assumed (to be) 
prime. Thus, the (set of) prime numbers A, B, C, G, 
(which is) more numerous than the assigned multitude 
(of prime numbers), A, B, C, has been found. (Which is) 
the very thing it was required to show. 



xa'. 

'Edv dpxioi dpidjioi ottoooiouv auvxe'dwaiv, 6 oXoc; 
dpxioc; eaxiv. 

A B T A E 

i 1 1 1 1 

Euyxeicxdwaav yap dpxioi dpid^tol OKoaoiouv oi AB, 
BT, TA, AE' Xeyco, Sxi oXo<; 6 AE dpxioc; eaxiv. 

'Euel yap exaaxoc xcov AB, BT, TA, AE dpxioc; eaxiv, 
exei ^epo<; f^tair toaxe xal oXoc 6 AE e/ei [lepoc, f^iou. 
dpxioc; 8e dpid^toc; eaxiv 6 8[)(a Biaipou^tevoc;- dpxioc dpa 
eaxiv 6 AE- ouep eSei SeT^ai. 



Proposition 21 

If any multitude whatsoever of even numbers is added 
together then the whole is even. 

A B C D E 

i 1 1 1 1 

For let any multitude whatsoever of even numbers, 
AB, BC, CD, DE, lie together. I say that the whole, 
AE, is even. 

For since everyone of AB, BC, CD, DE is even, it 
has a half part [Def. 7.6]. And hence the whole AE has 
a half part. And an even number is one (which can be) 
divided in half [Def. 7.6]. Thus, AE is even. (Which is) 



271 



ETOIXEIfiN fl'. 



ELEMENTS BOOK 9 



x(3'. 

'Eav Tcepiaaol apid^oi oTcoaoioOv auvxcdwaiv, to 8e 
TiXfj'dot; auxov apxiov fj, 6 oXoc; dpxioc; eaxai. 

A B T A E 

i 1 1 1 1 

Suyxeia'dcooav yap Tcepiaaol apid^ol 6aoi8r)icoxo0v 
dpxioi to TcXfjdoc: ol AB, Br, EA, AE- Xeyw, on oXoc; 
6 AE dpxioc; eaxiv. 

'EtccI yap exaaxoc; xwv AB, Br, TA, AE Tcepixxoc; eaxiv, 
dcpaipcMarjc; ^iovd8oc; dtp' exdaxou exaaxoc; xwv Xoitcov 
dpxioc; eaxar Saxe xal 6 auyxei^ievoc; e<; auxwv dpxioc; 
eaxai. eaxi 8e xal xo TtXfj'doc; xwv ^.ovdBwv apxiov. xal 
oXoc; dpa 6 AE dpxioc; eaxiv oTcep e8ei SeTc;ai. 



the very thing it was required to show. 

Proposition 22 

If any multitude whatsoever of odd numbers is added 
together, and the multitude of them is even, then the 
whole will be even. 

A B C D E 

i 1 1 1 1 

For let any even multitude whatsoever of odd num- 
bers, AB, BC, CD, DE, lie together. I say that the 
whole, AE, is even. 

For since everyone of AB, BC, CD, DE is odd then, a 
unit being subtracted from each, everyone of the remain- 
ders will be (made) even [Def. 7.7]. And hence the sum 
of them will be even [Prop. 9.21]. And the multitude 
of the units is even. Thus, the whole AE is also even 
[Prop. 9.21]. (Which is) the very thing it was required to 
show. 



xy'. 

'Eav Tiepiaaol dpid^iol OTcoaoioOv auvxrdwaiv, xo 8s 
TiXfj'dot; aOxov Tcepiaaov fj, xal 6 oXoc; icepiaaoc; eaxai. 

A B T E A 

i 1 — i 1 — i 

Xuyxeia'dwaav yap OTcoaoioOv Tiepiaaol dpid^ioi, Sv xo 
TcXrji9oc; Tcepiaaov eaxw, ol AB, Br, EA- Xeyw, oxi xal oXoc; 
6 AA icepiaaoc; eaxiv. 

Acpr]pr ! |a'do dico xoO IA ^tovdc; f\ AE- Xoitcoc; dpa 6 TE 
dpxioc; eaxiv. eaxi 8e xal 6 FA dpxioc;- xal oXoc; dpa 6 AE 
dpxioc; eaxiv. xai eaxi (lovac; r\ AE. Tcepiaaoc; dpa eaxiv 6 
AA- oTiep e8ei SeT^ai. 



Proposition 23 

If any multitude whatsoever of odd numbers is added 
together, and the multitude of them is odd, then the 
whole will also be odd. 

ABC ED 

i 1 1 1 — i 

For let any multitude whatsoever of odd numbers, 
AB, BC, CD, lie together, and let the multitude of them 
be odd. I say that the whole, AD, is also odd. 

For let the unit DE have been subtracted from CD. 
The remainder CE is thus even [Def. 7.7]. And CA 
is also even [Prop. 9.22]. Thus, the whole AE is also 
even [Prop. 9.21]. And DE is a unit. Thus, AD is odd 
[Def. 7.7]. (Which is) the very thing it was required to 
show. 



x5'. 

'Eav dTto dpxbu dpi%ou dpxioc; d(paipei5fj, 6 Xoitcoc; 
dpxioc; eaxai. 

a r b 

i 1 1 

Atco yap dpxiou xou AB dpxioc; dcprprja'dco 6 BE Xeyw, 
oxi 6 Xoitcoc; 6 EA dpxioc; eaxiv. 

'Etcci yap 6 AB dpxioc; eaxiv, e^ei ^tepoc; y^iau. 8ia ^d 
auxd Br| xal 6 BT exei ^epoc; r^iatr Saxe xal Xoitcoc; [6 FA 
exei [iepoc; r^iau] dpxioc; [dpa] eaxiv 6 AT- oTiep e8ei 8eTi;ai. 



Proposition 24 

If an even (number) is subtracted from an(other) even 
number then the remainder will be even. 

A C B 



For let the even (number) BC have been subtracted 
from the even number AB. I say that the remainder CA 
is even. 

For since AB is even, it has a half part [Def. 7.6]. So, 
for the same (reasons), BC also has a half part. And 
hence the remainder [CA has a half part]. [Thus,] AC is 
even. (Which is) the very thing it was required to show. 



272 



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xe'. 

'Edv duo dpxiou dpidfiou Ttepiaaoc; dcpaipe'drj, 6 Xoitcoc; 
Ttepiaaoc; eaxai. 

A r A B 

I 1 1 1 

Atco yap dpxiou xou AB Ttepiaaoc; dcpr)pf|a , d« 6 BE 
Xeyco, oxi 6 Xomoc; 6 EA Ttepiaaoc; eaxiv. 

Acpr)pr]a'do jap and xou Br \lovolz f] EA- 6 AB dpa 
apxioc eaxiv. eaxi 8e xal 6 AB apxioc xal XoiTtoc; dpa 6 
AA apxioc eaxiv. xai eaxi (iovdc f] EA- 6 EA dpa Ttepiaaoc 
eaxiv oitep e8ei Belial. 



Xf'. 

'Eav duo Ttepiaaou dpid^iou Ttepiaaoc dcpaipe'drj, 6 XoiTtoc 
apxioc eaxai. 

A r A B 

I 1 1 1 

Atco yap Ttepiaaou xou AB Ttepiaaoc dcpr)pr]a , do 6 BE 
Xey«, oxi 6 Xomoc 6 EA apxioc eaxiv. 

Tkei yap 6 AB Ttepiaaoc eaxiv, dcpr)pf|a , dco [lovac, f] BA- 
XoiTtoc dpa 6 AA apxioc eaxiv. Bid xd auxd 5r] xal 6 EA 
apxioc eaxiv Saxe xai XoiTtoc 6 EA apxioc eaxiv oTtep e8ei 
SeT^ai. 



'Edv aTto Ttepiaaou dpi'd^iou apxioc dcpaipeiJrj, 6 XoiTtoc 
Ttepiaaoc eaxai. 

A A r B 

I 1 1 1 

Atco yap Ttepiaaou xou AB apxioc dcpr]pr]a , d« 6 BE 

Xeyo, oxi 6 Xomoc 6 EA Ttepiaaoc eaxiv. 

Acpr)pr]a , dw [yap] [iovdc f) AA- 6 AB dpa apxioc; eaxiv. 

eaxi 8e xal 6 Br apxioc;- xal Xomoc dpa 6 EA apxioc; eaxiv. 

Ttepiaaoc dpa 6 EA- OTtep e8ei 5eTc;ai. 



XT]'. 

'Eav Ttepiaaoc dpid^toc apxiov TtoXXaTtXaaidaac Ttoirj 
xiva, 6 yevojievoc apxioc; eaxai. 



Proposition 25 

If an odd (number) is subtracted from an even num- 
ber then the remainder will be odd. 

A CD B 



For let the odd (number) BC have been subtracted 
from the even number AB. I say that the remainder CA 
is odd. 

For let the unit CD have been subtracted from BC. 
DB is thus even [Def. 7.7]. And AB is also even. And 
thus the remainder AD is even [Prop. 9.24]. And CD is 
a unit. Thus, CA is odd [Def. 7.7]. (Which is) the very 
thing it was required to show. 

Proposition 26 

If an odd (number) is subtracted from an odd number 
then the remainder will be even. 

A C D B 



For let the odd (number) BC have been subtracted 
from the odd (number) AB. I say that the remainder CA 
is even. 

For since AB is odd, let the unit BD have been 
subtracted (from it). Thus, the remainder AD is even 
[Def. 7.7]. So, for the same (reasons), CD is also 
even. And hence the remainder CA is even [Prop. 9.24]. 
(Which is) the very thing it was required to show. 

Proposition 27 

If an even (number) is subtracted from an odd num- 
ber then the remainder will be odd. 

AD C B 

i 1 1 1 

For let the even (number) BC have been subtracted 
from the odd (number) AB. I say that the remainder CA 
is odd. 

[For] let the unit AD have been subtracted (from AB) . 
DB is thus even [Def. 7.7]. And BC is also even. Thus, 
the remainder CD is also even [Prop. 9.24]. CA (is) thus 
odd [Def. 7.7]. (Which is) the very thing it was required 
to show. 

Proposition 28 

If an odd number makes some (number by) multiply- 
ing an even (number) then the created (number) will be 
even. 



273 



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A i 1 A^ 



B ' 1 

r i 1 

Hspiaooz ydp api%6<; 6 A dpxiov xov B TtoXXa- 
TtXaaidaac; tov T itoishw Xeyio, oxi 6 T apxioc; eaxiv. 

'End yap 6 A tov B TroXXaTtXaaidaac; tov T kstioi/jxev, 
6 T dpa auyxeixai ex xoaouxtov lacov ifi B, oaai rialv sv 
xG A (iovdSeg. xai eaxiv 6 B apxioc 6 T apa auyxeixai 
z\ dpxicov. edv 6e dpxioi dpidjjiol onoaoiouv auvxe-dGaiv, 6 
oXoc; apxioc; eaxiv. apxiog apa eaxiv 6 T- onep eBei 5eT<;ai. 



X0'. 

'Edv rcepiaaoc; dpi'd^oc; Ttepiaaov dpid^tov TtoXXauXaaidc;- 
ac, Ttoifj Tiva, 6 yev6[ievo<; Ttepiaaoc; eaxai. 

A' 1 

B i 1 

r i 1 

ilepiaaoc; yap dpid^ioc; 6 A Ttepiaaov tov B uoXXa- 
jcXaaidaac; tov T Tioieixw Xeyco, oxi 6 T Ttepiaaoc; eaxiv. 

'Ercel yap 6 A tov B TioXXajcXaaidaac; tov T TieTcoirjxev, 
6 T apa auyxeixai ex xoaouxov laov xG B, oaai eialv ev iu 
A ^ovdBec;. xa( eotiv exdxepoc; xGv A, B Ttepiaaoc 6 T apa 
auyxeixai ex TtepiaaGv dpid^iGv, Gv to TiXfj-doi; Ttepiaaov 
eaxiv. uoxe 6 T Ttepiaaoc; eaxiv oTtep eSei Belial. 



X'. 

'Edv Ttepiaaoc; dpid^oc; dpTiov dpnf^ov ^Expfj, xal tov 
r^iauv auxou jiexprjoei. 

A i 1 

B i 1 

r i 1 

Ilepiaaoc; yap dprd^ioc; 6 A dpxiov xov B ^.sxpeixw Xeyio, 
oxi xal xov y](itauv auxou ^exprpei. 

'EtccI yap 6 A xov B [icxpel, ^texpeixM auxov xaxd xov 
F' Xeyco, oxi 6 T oux eaxi Ttepiaaoc;. el yap Suvaxov, eaxa>. 
xal eTtel 6 A xov B [iexpeT xaxd xov T, 6 A apa xov T 
TtoXXaTtXaaidaac; xov B TteTtofyxev. 6 B apa auyxeixai ex 
TtepiaaGv dpn9[iGv, Gv xo TtXTj-doc; Ttepiaaov eaxiv. 6 B apa 



Bi 1 

Ci 1 

For let the odd number A make C (by) multiplying 
the even (number) B. I say that C is even. 

For since A has made C (by) multiplying B, C is thus 
composed out of so many (magnitudes) equal to B, as 
many as (there) are units in A [Def. 7.15]. And B is 
even. Thus, C is composed out of even (numbers) . And 
if any multitude whatsoever of even numbers is added 
together then the whole is even [Prop. 9.21]. Thus, C is 
even. (Which is) the very thing it was required to show. 

Proposition 29 

If an odd number makes some (number by) multiply- 
ing an odd (number) then the created (number) will be 
odd. 

A' ' 

B' 1 

C' 1 

For let the odd number A make C (by) multiplying 
the odd (number) B. I say that C is odd. 

For since A has made C (by) multiplying B, C is thus 
composed out of so many (magnitudes) equal to B, as 
many as (there) are units in A [Def. 7.15]. And each 
of A, B is odd. Thus, C is composed out of odd (num- 
bers), (and) the multitude of them is odd. Hence C is odd 
[Prop. 9.23]. (Which is) the very thing it was required to 
show. 

Proposition 30 

If an odd number measures an even number then it 
will also measure (one) half of it. 

A i 1 

B' 1 

Ci 1 

For let the odd number A measure the even (number) 

B. I say that (A) will also measure (one) half of (B). 
For since A measures B, let it measure it according to 

C. I say that C is not odd. For, if possible, let it be (odd). 
And since A measures B according to C, A has thus made 
B (by) multiplying C. Thus, B is composed out of odd 
numbers, (and) the multitude of them is odd. B is thus 



274 



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ELEMENTS BOOK 9 



Ttepiaaoc eaxiv onep axorcov UTtoxeixai yap dpxioc;. oux 
apa 6 r Ttepiaaoc; eaxiv dpxioc; apa eaxiv 6 T. waxe 6 A 
xov B ^texpeT dpxidxic;. Bid 8rj xouxo xal xov fj^iauv auxou 
[Lejpfiaei- oiiep e8ei BeT^ai. 

Xa'. 

'Edv icepiaaoc; dpid^toc; upoc; xiva dpid[i6v rcpaixoc; rj, xal 
npoQ xov 8i7iXaa(ova auxou Kpwxoc eaxai. 

A' 1 

B ' 1 

r i 1 

A' ' 

TLzpiaaoc, yap dpiif)[i6<; 6 A Ttpoc; xiva dpid^iov xov B 
npoxoc; eaxw, xou 8e B SmXaoiwv eaxto 6 E Xeyoj, 0X1 ° A 
[xal] upoc; xov T rcpcoxoc; eaxiv. 

Ei yap \ir\ eiaiv [ol A, T] Ttpfixoi, jiexpfpei xic; auxoug 
dpi-f^oc;. jiexpeixw, xal eaxto 6 A. xal eaxiv 6 A Ttepiaaoc;- 
Ttepiaaoc; apa xal 6 A. xal end 6 A Ttepiaaoc; uv xov T 
[icxpeT, xa[ eaxiv 6 T apxioc;, xal xov rj[iiauv apa xou T 
(jiexpiqaei [6 A], xou 8e T fj^uau eaxiv 6 B- 6 A apa xov 
B ^lexpeT. ^texpeT Be xal xov A. 6 A apa xou? A, B ^texpeT 
Ttpwxouc; ovxac; Ttpoc; dXXrjXouc oTtep eaxlv dSuvaxov. oux 
apa 6 A Ttpoc; xov T TtpGxoc; oux eaxiv. oi A, T apa TtpGxoi 
Ttpoc; dXXr|Xou<; eiaiv OTtep eBei 8eTc;ai. 



X(3'. 

Tcov duo 8ua8oc; 8mXaaia^ou.ev«v dpi-djiwv exaaxoc; 
dpxidxic; dpxioc; eaxi [iovov. 

A' ' 

B i 1 

r i 1 

A' ' 

Atco yap 8uaBoc; xfjc; A 8eBiTtXaaidaiL>Maav baoihr]- 
noxouv dpi^ol oi B, T, A- Xeyw, oxi oi B, T, A dpxidxic; 
dpxioi eiai [iovov. 

"Oxi [lev ouv exaaxoc; [xwv B, T, A] dpxidxic; dpxioc; 
eaxiv, cpavepov aTto ydp Bud8oc; eaxi BiTtXaaiaai&eic;. Xeyw, 
oxi xal [iovov. exxeicdw ydp [iovdc;. enel ouv aTto [iovdBoc; 
ojioaoiouv dpid[iol e^fjg dvdXoyov eiaiv, 6 8e [iexd xrjv 
[iovd8a 6 A upwxoc; eaxiv, 6 [ieyiaxoc; xfiv A, B, T, A 6 



odd [Prop. 9.23]. The very thing (is) absurd. For (B) 
was assumed (to be) even. Thus, C is not odd. Thus, C 
is even. Hence, A measures B an even number of times. 
So, on account of this, (A) will also measure (one) half 
of (£>). (Which is) the very thing it was required to show. 

Proposition 31 

If an odd number is prime to some number then it will 
also be prime to its double. 

A' 1 

B 1 

Ci ' 

D' 1 

For let the odd number A be prime to some number 
B. And let C be double B. I say that A is [also] prime to 
C. 

For if [A and C] are not prime (to one another) then 
some number will measure them. Let it measure (them), 
and let it be D. And A is odd. Thus, D (is) also odd. 
And since D, which is odd, measures C, and C is even, 
[D] will thus also measure half of C [Prop. 9.30]. And B 
is half of C. Thus, D measures B. And it also measures 
A. Thus, D measures (both) A and B, (despite) them 
being prime to one another. The very thing is impossible. 
Thus, A is not unprime to C. Thus, A and C are prime to 
one another. (Which is) the very thing it was required to 
show. 

Proposition 32 

Each of the numbers (which is continually) doubled, 
(starting) from a dyad, is an even-times-even (number) 
only. 

A' ' 

B 1 

Ci ' 

D' 1 

For let any multitude of numbers whatsoever, B, C, 
D, have been (continually) doubled, (starting) from the 
dyad A. I say that B, C, D are even-times-even (num- 
bers) only. 

In fact, (it is) clear that each [of B, C, D] is an 
even-times-even (number) . For it is doubled from a dyad 
[Def. 7.8]. I also say that (they are even-times-even num- 
bers) only. For let a unit be laid down. Therefore, since 



275 



ETOIXEIfiN fl'. 



ELEMENTS BOOK 9 



A On' ouSevoc; aXXou [j.exprj'driaexai napei; ^wv A, B, T. xod 
eaxiv exaaxoc; xwv A, B, T apxioc 6 A apa dpxidxic; apxioc; 
eaxi [lovov. b[io'u^z 8f) 8e(c;o[iev, oxi [xal] exdxepoc; xwv B, 
T dtpxidxic; apxiog eaxi [xovov onep eSei Sei<;ai. 



Xy'. 

'Eav dpid^ioc; xov fjjiiauv e/r) nepiaaov, dpxidxic; ne- 
piaaoc; eaxi [iovov. 

A' 1 

'Apvdjidc; yap 6 A xov fjuiauv 'tyzTU> nepiaaov Xeyto, oxi 
6 A dpxidxic; nepiaaoc; eaxi [iovov. 

"Oxi [lev ouv dpxidxic; nepiaaoc; eaxiv, cpavepov 6 yap 
r^iauc; auxou nepioaoc &v [iexpeT aOxov dpxidxic;, Xeyco Sr|, 
oxi xal \iovov. eE yap eaxai 6 A xal dpxidxic; apxioc;, [ls- 
xp/jOiqaexai uno dpxiou xaxd apxiov dprd^tov uoxe xal 6 
f\\x\.av>c, auxou jiexp/ji&iqaexai uno dpxiou dpidjiou nepiaaoc; 
uv onep eaxiv axonov. 6 A apa dpxidxic; nepioaoc; eaxi 
[iovov onep eSei 5el<;ai. 



X8'. 

'Eav dpii9|ji6c; \JX\ie xwv and 8ud8o<; BinXaaiai^o^evwv fj, 
y.r]TS xov r^iauv z^T] nepiaaov, dpxidxic; xe apxioc; eaxi xal 
dpxidxic; nepiaaoc;. 

A' 1 

Api-djjioc; yap 6 A [ir|xe xGv dno 8ud8oc; 8mXaaia£o[iivov 
eaxw [lifts xov fjuiauv tyixu nepiaaov Xeyw, oxi 6 A 
dpxidxic; xe eaxiv apxioc; xal dpxidxic; nepiaaoc- 

"Oxi [lev ouv 6 A dpxidxic; eaxiv apxioc;, cpavepov xov 
yap rjuiauv oux exei nepiaaov. Xey« 8rj, oxi xal dpxidxic; ne- 
piaaoc eaxiv. eav yap xov A xeuvwuev 8()(a xal xov fjuiauv 
auxou 8[)(a xal xouxo del noifiuev, xaxavxrjaouev zic, xiva 
dpiduov nepiaaov, oc uexpf|aei xov A xaxd apxiov dpiduov. 
ei yap ou, xaxavxyjaouev eic 8ud8a, xal eaxai 6 A xwv dno 
8ud8oc 8mXaaia£ouev«v onep oux unoxeixai. waxe 6 A 
dpxidxic; nepiaaov eaxiv. e8e[)fdr] 8e xal dpxidxic; apxioc;. 6 
A apa dpxidxic; xe apxioc; eaxi xal dpxidxic; nepiaaoc onep 
e8ei 8elc;ai. 



any multitude of numbers whatsoever are continuously 
proportional, starting from a unit, and the (number) A af- 
ter the unit is prime, the greatest of A, B, C, D, (namely) 
D, will not be measured by any other (numbers) except 
A, B, C [Prop. 9.13]. And each of A, B, C is even. Thus, 
D is an even-time-even (number) only [Def. 7.8]. So, 
similarly, we can show that each of B, C is [also] an even- 
time-even (number) only. (Which is) the very thing it was 
required to show. 

Proposition 33 

If a number has an odd half then it is an even-time- 
odd (number) only. 

A i 1 

For let the number A have an odd half. I say that A is 
an even-times-odd (number) only. 

In fact, (it is) clear that (A) is an even-times-odd 
(number). For its half, being odd, measures it an even 
number of times [Def. 7.9]. So I also say that (it is 
an even-times-odd number) only. For if A is also an 
even-times-even (number) then it will be measured by an 
even (number) according to an even number [Def. 7.8]. 
Hence, its half will also be measured by an even number, 
(despite) being odd. The very thing is absurd. Thus, A 
is an even-times-odd (number) only. (Which is) the very 
thing it was required to show. 

Proposition 34 

If a number is neither (one) of the (numbers) doubled 
from a dyad, nor has an odd half, then it is (both) an 
even-times-even and an even-times-odd (number) . 

A i 1 

For let the number A neither be (one) of the (num- 
bers) doubled from a dyad, nor let it have an odd half. 
I say that A is (both) an even-times-even and an even- 
times-odd (number). 

In fact, (it is) clear that A is an even-times-even (num- 
ber) [Def. 7.8]. For it does not have an odd half. So I 
say that it is also an even-times-odd (number) . For if we 
cut A in half, and (then cut) its half in half, and we do 
this continually, then we will arrive at some odd num- 
ber which will measure A according to an even number. 
For if not, we will arrive at a dyad, and A will be (one) 
of the (numbers) doubled from a dyad. The very oppo- 
site thing (was) assumed. Hence, A is an even-times-odd 
(number) [Def. 7.9]. And it was also shown (to be) an 
even-times-even (number). Thus, A is (both) an even- 
times-even and an even-times-odd (number) . (Which is) 



276 



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ELEMENTS BOOK 9 



Xe'. 

'Edv fiaiv 6aoiSr)7ioxouv apidfiol kc,r\c, dvdXoyov, dcpai- 
pcdfiai 8e duo xe xou Seuxepou xal tou ea)(dxou laoi iu 
Ttpcbxw, eaxai ox f) xou 8euxepou UTiepo/r) Ttpoc; xov Ttpfixov, 
ouxcoe; f] tou ea/dxou (mepoxr) Ttpoc; xouc; Tipo eauxou 
ndvxac;. 

A' 1 

b h r 

I 1 1 



the very thing it was required to show. 

Proposition 35 f 

If there is any multitude whatsoever of continually 
proportional numbers, and (numbers) equal to the first 
are subtracted from (both) the second and the last, then 
as the excess of the second (number is) to the first, so the 
excess of the last will be to (the sum of) all those (num- 
bers) before it. 

A' ' 

B G C 

i 1 — i 



A 1 

E A K Z 

i — i — i — i 1 

'Eaxcoaav oTioaoiSrjTtoxouv dpi/d^io! dvdXoyov oi A, 
Br, A, EZ dcp^o^evoi aTto eXaxiaxou xou A, xdi dcpr)pr]C7d« 
dno xou Br xdi xou EZ xcb A loot; exdxepoc; xfiv BH, Z0- 
Xeyw, oxi eaxiv cbe; 6 BT npoc, xov A, ouxioc; 6 E0 Ttpoc; 
xouc; A, Br, A. 

Keicdco y&P t<P Br Xaoc, 6 ZK, xcp 8e A Xaoc, 6 ZA. 
xal inei 6 ZK xfi Br Xaoc, eaxiv, Sv 6 Z0 iu BH lao? eaxiv, 
Xoiitoc; dpa 6 0K Xouuo xQ HT eaxiv laoc;. xdi eitei eaxiv ox 
6 EZ Ttpoc; xov A, ouxox 6 A Ttpoc; xov Br xdi 6 Br Ttpoc; 
xov A, iao<; 8e 6 ^tev A xw ZA, 6 8e Br xw ZK, 6 8e A iw 
ZO, eaxiv dpa ox 6 EZ Ttpoc; xov ZA, ouxox 6 AZ Ttpoc; xov 
ZK xdi 6 ZK Ttpoc; xov Z0. SieXovxi, ox 6 EA Ttpoc; xov AZ, 
ouxox 6 AK Ttpoc; xov ZK xdi 6 KG Ttpoc; xov Z0. eaxiv dpa 
xal ox zXc, xwv r)you[ievojv npoc eva xwv eTtojievojv, ouxox 
aTtavxec; oi rjyou^ievoi npoc aTtavxac; xouc; CTto|ievou<;- eaxiv 
dpa ox 6 KG Ttpoc; xov Z0, ouxox oi EA, AK, KG Ttpoc; 
xou? AZ, ZK, 6Z. iaoc 8e 6 yttv K9 xo> TH, 6 Se Z0 xo> 
A, oi Se AZ, ZK, 6Z xol? A, Br, A- eaxiv dpa ox 6 TH 
Ttpoc; xov A, ouxox 6 E0 Ttpoc; xoix A, Br, A. eaxiv dpa 
ox f] xou 8euxepou UTiepoxr] npoc xov Ttpwxov, ouxox f\ xou 
ea)(dxou UTtepoxr] Ttpoc; xoix Ttpo eauxou Ttdvxac oitep e8ei 
8eT^ai. 

t This proposition allows us to sum a geometric series of the form a, 
(ar- a)/a = (ar n - a)/S„. Hence, 5„ = a (r n - l)/(r - 1). 

X<r'. 

'Edv aTto [iovd8oc; onoaoiouv dpidjiol e^fjc; exxeiDwaiv ev 
xfj BiTtXaaiovi dvaXoyia, eox ou 6 au^iTtac; auvxe-deic; Ttpfiixoc; 
yevrjxai, xal 6 au^iTtac; era xov ea)(axov TtoXXaTtXaaiaaiJelc; 



D 1 

E L K H F 

i 1 1 1 1 

Let A, BC, D, EF be any multitude whatsoever of 
continuously proportional numbers, beginning from the 
least A. And let BG and FH, each equal to A, have been 
subtracted from BC and EF (respectively). I say that as 
GC is to A, so EH is to A, BC, D. 

For let FK be made equal to BC, and FL to D. And 
since FK is equal to BC, of which FH is equal to BG, 
the remainder 7? K is thus equal to the remainder GC. 
And since as EF is to £>, so D (is) to BC, and BC to 
A [Prop. 7.13], and D (is) equal to FL, and BC to Bif, 
and A to FiJ, thus as EF is to FL, so LF (is) to FJf, and 
FK to Fff. By separation, as EL (is) to LF, so £AT (is) 
to FK, and FiJ to FiJ [Props. 7.11, 7.13]. And thus as 
one of the leading (numbers) is to one of the following, 
so (the sum of) all of the leading (numbers is) to (the 
sum of) all of the following [Prop. 7.12]. Thus, as KH 
is to FH, so EL, LK, KH (are) to LF, FK, HF. And 
KH (is) equal to CG, and FH to A, and LF, FK, HF 
to D, BC, A. Thus, as CG is to A, so EH (is) to D, 
BC, A. Thus, as the excess of the second (number) is to 
the first, so the excess of the last (is) to (the sum of) all 
those (numbers) before it. (Which is) the very thing it 
was required to show. 

ar, ar 2 , ar 3 , ■ ■ ■ ar"^ 1 . According to Euclid, the sum S n satisfies 



Proposition 36 f 

If any multitude whatsoever of numbers is set out con- 
tinuously in a double proportion, (starting) from a unit, 
until the whole sum added together becomes prime, and 



277 



ETOIXEIfiN fl'. 



ELEMENTS BOOK 9 



Ttoifj xiva, 6 yev6^ievo<; xeXeioc; eaxai. 

Atco yap ^iovd8o<; exxeia-dwaav 6ooi8t)tioto0v dpi-d^- 
ol ev xfj SmXaaiovi dvaXoyia, ewe; ou 6 aujjmac; auvxei&eic; 
7tp£>xo<; yevrjxai, oi A, B, T, A, xai x£> aujiiravxi iao? eaxw 
6 E, xal 6 E xov A TtoXXaTtXaaidaac; xov ZH Ttoieixco. Xeyco, 
oxi 6 ZH xeXeioc eaxiv. 

B i — i 

r i ' 

A | 1 



the sum multiplied into the last (number) makes some 
(number), then the (number so) created will be perfect. 

For let any multitude of numbers, A, B, C, D, be set 
out (continuouly) in a double proportion, until the whole 
sum added together is made prime. And let E be equal to 
the sum. And let E make FG (by) multiplying D. I say 
that FG is a perfect (number) . 

B 1 

C' ' 

D 1 



E' — ' 
N K 

A 1 

Mi 1 

Z H H 

I 1 1 

O ' 

n ' 

"Oaoi ydp eiaiv oi A, B, T, A xfi> rcXrydei, xoaouxoi duo 
xou E £iXf](pTf)«aav ev xfj SmXaaiovi dvaXoyia oi E, 9K, A, 
M - 8i' i'aou dpa laxlv &>c, 6 A Ttp6<; xov A, ouxcoc; 6 E npoc; 
xov M. 6 dpa ex x«v E, A i'aoc; eaxi x£> ex xwv A, M. xai 
eaxiv 6 ex xQv E, A 6 ZH- xai 6 ex xuv A, M dpa eaxiv 6 
ZH. 6 A dpa xov M TtoXXaTtXaaidaac; xov ZH TteTtoirjxev 6 
M dpa xov ZH ^.expeT xaxa xa<; ev x£> A ^.ovdBac;. xai eaxi 
Suae; 6 A- 8iTtXdaio<; dpa eaxiv 6 ZH xou M. eiai 8e xai oi M, 
A, 0K, E e^rjej SmXdoioi dXXf|X«v oi E, 0K, A, M, ZH dpa 
e^r\Q dvdXoyov eiaiv ev xfj BiTtXaaiovi dvaXoyia. dcpr)pf|a , d« 
8rj duo xou Beuxepou xou 6K xai xou eo^axou xou ZH xu 
Tipcoxw xw E i'aoc; exdxepo<; xwv 9N, ZS - eaxiv dpa d>c f\ 
xou Seuxepou dprd|iou UKepo^f] 7ip6<; xov Tipfixov, ouxoc; f) 
xou ea)(dxou UTiepo)(r) 7ipo<; xouc; Ttpo eauxou Ttdvxac;. eaxiv 
dpa &><; 6 NK 7ip6<; xov E, ouxwc; 6 SH 7ip6<; xouc; M, A, 
KG, E. xai eaxiv 6 NK i'aoc; x£> E - xai 6 SH dpa Taoc; eaxi 
xdic M, A, 8K, E. eaxi 8e xai 6 ZS xw E iaoz, 6 8e E 
xou; A, B, r, A xai xfj ^tovd8i. oXoc; dpa 6 ZH iaoc; eaxi 
xou; xe E, 9K, A, M xai xou; A, B, T, A xai xfj ^iovd8i- 
xai ^lexpeTxai bn auxwv. Xeyw, oxi xai 6 ZH viz' ou8evo<; 
dXXou ^exprj-driaexai nape? xwv A, B, T, A, E, 9K, A, M 
xai xfjc ^.ovdBoc;. ei yap 8uvax6v, ^.expeixw xic; xov ZH 6 
O, xai 6 O [LrpeVi xQv A, B, T, A, E, 9K, A, M eaxw 6 
auxocj. xai oadxu; 6 O xov ZH jiexpel, xoaauxai ^iovd8e<; 



Ei ' 

H N K 

I 1 1 

L i 1 

Mi ' 

F O G 

P i 1 

Q 

For as many as is the multitude of A, B, C, D, let so 
many (numbers), E, HK, L, M, have been taken in a 
double proportion, (starting) from E. Thus, via equal- 
ity, as A is to D, so E (is) to M [Prop. 7.14]. Thus, the 
(number created) from (multiplying) E, D is equal to the 
(number created) from (multiplying) A, M. And FG is 
the (number created) from (multiplying) E, D. Thus, 
FG is also the (number created) from (multiplying) A, 
M [Prop. 7.19]. Thus, A has made FG (by) multiplying 
M. Thus, M measures FG according to the units in A. 
And A is a dyad. Thus, FG is double M. And M, L, 
HK, E are also continuously double one another. Thus, 
E, HK, L, M, FG are continuously proportional in a 
double proportion. So let HN and FO, each equal to the 
first (number) E, have been subtracted from the second 
(number) HK and the last FG (respectively). Thus, as 
the excess of the second number is to the first, so the ex- 
cess of the last (is) to (the sum of) all those (numbers) 
before it [Prop. 9.35]. Thus, as NK is to E, so OG (is) 
to M, L, KH, E. And NK is equal to E. And thus OG 
is equal to M, L, HK, E. And FO is also equal to E, 
and E to A, B, C, D, and a unit. Thus, the whole of FG 
is equal to E, HK, L, M, and A, B, C, D, and a unit. 
And it is measured by them. I also say that FG will be 



278 



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eaxtoaav ev iu II ■ 6 II apa xov O rcoXXaTiXaaidaat; xov ZH 
Tten:o[r)xev. dXXd [jnrjv xai 6 E xov A TtoXXaTtXaaidaac; xov 
ZH 7CE7ioir]xev eaxiv apa cbc; 6 E Tipoc xov II, 6 O Tipoc; xov 
A. xal etcel duo ^lovdBoc; e<;rj<; dvdXoyov eiaiv ol A, B, T, 
A, 6 A apa On' ouSevog aXXou dpi-djioO jj.sxprj'diqaexai nape? 
xcov A, B, T. xal UTioxeixai 6 O ouSevi iSv A, B, T 6 auxoc 
oux apa ^.expf|aei 6 O xov A. dXX' «<; 6 O 7tp6<; xov A, 6 
E 7ip6<; xov II- ou8e 6 E apa xov II ^.expel. xa[ eaxiv 6 E 
7ipoxo<;- n&z 8e Ttpwxoc; dpi-djaot; npbc, anavxa, 6v \ir\ jiexpe'i, 
7tp£>x6<; [eaxiv]. oi E, II apa npGxoi Kpog dXXr|Xou<; eiaiv. oi 
8e npwxoi xal eXd)(iaxoi, oi 8e eXd)(iaxoi ^.expouai xou<; xov 
auxov Xoyov ey^ovxac; iadxic; o xe rjyounevoi; xov rjyou^ievov 
xal 6 £iz6[ievoci xov CTtojievov xa[ eaxiv «<; 6 E Ttp6<; xov n, 
6 O Tip6<; xov A. iadxi<; apa 6 E xov O (jiexpeT xal 6 II xov 
A. 6 8e A Cm' ouBevoc dXXou ^expeTxai nape? xov A, B, T- 
6 n apa evl xov A, B, T eaxiv 6 auxoc;. eaxco twBo auxoc;. 
xal oaoi eiaiv oi B, T, A xw TtXrydei xoaoOxoi eiXTjcpiJoaav 
aTio xou E oi E, 6K, A. xai eiaiv oi E, 9K, A toXq B, T, A 
ev xw auxw Xoyw 8i' taou apa eaxiv w?6B Kpoc; xov A, 6 
E 7ip6<; xov A. 6 apa ex xwv B, A Taoc eaxl xo ex xQv A, 
E- dXX' 6 ex xwv A, E iao<; eaxi x« ex xwv II, O- xal 6 ex 
xwv n, O apa Xaoc, eaxl iw ex xwv B, A. eaxiv apa w<; 6 II 
7ip6<; xov B, 6 A 7ip6<; xov O. xai eaxiv 6 II xw B 6 auxoc - 
xal 6 A apa xw O eaxiv 6 auxoc ojiep d86vaxov 6 yap O 
UTioxeixai jir]8evl xwv exxeijievcov 6 auxoc oux apa xov ZH 
\LZxpr\oz\. tiz dpn!)^6<; Kape? xCSv A, B, T, A, E, 9K, A, M 
xal xfj? jiovd8o<;. xal eBeix/] 6 ZH xoI<; A, B, T, A, E, 0K, 
A, M xal xfj jiovdBi Xaoc,. xeXeioi; 8e dpiif)[i6<; eaxiv 6 xou; 
eauxou [lepeaiv Xaoz wv xeXeioc apa eaxiv 6 ZH- oTiep e8ei 
8el^ai- 



measured by no other (numbers) except A, B, C, D, E, 
HK, L, M, and a unit. For, if possible, let some (num- 
ber) P measure FG, and let P not be the same as any 
of A, B, C, D, E, HK, L, M. And as many times as P 
measures FG, so many units let there be in Q. Thus, Q 
has made FG (by) multiplying P. But, in fact, E has also 
made FG (by) multiplying D. Thus, as E is to Q, so P 
(is) to D [Prop. 7.19]. And since A, B, C, D are con- 
tinually proportional, (starting) from a unit, D will thus 
not be measured by any other numbers except A, B, C 
[Prop. 9.13]. And P was assumed not (to be) the same 
as any of A, B, C. Thus, P does not measure D. But, 
as P (is) to D, so E (is) to Q. Thus, E does not mea- 
sure Q either [Def. 7.20]. And E is a prime (number). 
And every prime number [is] prime to every (number) 
which it does not measure [Prop. 7.29]. Thus, E and Q 
are prime to one another. And (numbers) prime (to one 
another are) also the least (of those numbers having the 
same ratio as them) [Prop. 7.21], and the least (num- 
bers) measure those (numbers) having the same ratio as 
them an equal number of times, the leading (measuring) 
the leading, and the following the following [Prop. 7.20]. 
And as E is to Q, (so) P (is) to D. Thus, E measures P 
the same number of times as Q (measures) D. And D 
is not measured by any other (numbers) except A, B, C. 
Thus, Q is the same as one of A, B, C. Let it be the same 
as B. And as many as is the multitude of B, C, D, let so 
many (of the set out numbers) have been taken, (start- 
ing) from E, (namely) E, HK, L. And E, HK, L are in 
the same ratio as B, C, D. Thus, via equality, as B (is) 
to D, (so) E (is) to L [Prop. 7.14]. Thus, the (number 
created) from (multiplying) B, L is equal to the (num- 
ber created) from multiplying D, E [Prop. 7.19]. But, 
the (number created) from (multiplying) D, E is equal 
to the (number created) from (multiplying) Q, P. Thus, 
the (number created) from (multiplying) Q, P is equal 
to the (number created) from (multiplying) B, L. Thus, 
as Q is to B, (so) L (is) to P [Prop. 7.19]. And Q is the 
same as B. Thus, L is also the same as P. The very thing 
(is) impossible. For P was assumed not (to be) the same 
as any of the (numbers) set out. Thus, FG cannot be 
measured by any number except A, B, C, D, E, HK, L, 
M, and a unit. And FG was shown (to be) equal to (the 
sum of) A, B, C, D, E, HK, L, M, and a unit. And a 
perfect number is one which is equal to (the sum of) its 
own parts [Def. 7.22]. Thus, FG is a perfect (number). 
(Which is) the very thing it was required to show. 



t This proposition demonstrates that perfect numbers take the form 2 n 1 (2 n — 1) provided that 2 n — 1 is a prime number. The ancient Greeks 
knew of four perfect numbers: 6, 28, 496, and 8128, which correspond to n = 2, 3, 5, and 7, respectively. 



279 



280 



ELEMENTS BOOK 10 



Incommensurable Magnitudes^ 



tThe theory of incommensurable magntidues set out in this book is generally attributed to Theaetetus of Athens. In the footnotes throughout 
this book, k, k' , etc. stand for distinct ratios of positive integers. 



281 



ETOIXEIftN i'. 



ELEMENTS BOOK 10 



"Opoi. 

a'. Eujijiexpa jieye-dT] Xeyexai ™ T <?> auxG [xexpco fis- 
xpoujieva, dau^expa 56, 5v [ir)8ev evSexsxai xoivov [isxpov 
yevecrdai. 

P'. EMeTai SuvdjieL au^expoi ricnv, oxav xd an auxfiiv 
xexpdywva xw ocuxw X W P'<P ^ETpfjxai, dau^expoi 8e, oxav 
xou; dm' auxwv xexpaycovou; p]5ev £v8e)(r)xai x w p' ov xoivov 
^.fxpov yevecrdoa. 

y'. Touxwv Otioxei^ievwv Seixvuxai, oxi xrj Ttpoxrfteiar] 
su-Ma uudpxouaiv £015610(1 TtXyydei ameipoi au^iexpoi xe xod 
dau^expoi ai [ie\> [i^xei ^tovov, ai 8s xal Suvd^iei. xaXdcrdM 
ouv f] [iev Tipoxe-delaa eu-fMa prjxr], xal ai xauxr] au^expoi 
eixs [xr]X£i xal Suvd^iei eixe 5uvd^i£i \i6mov pr]xa(, ai 5e xauxr) 
dau[i^t£xpoi dXoyoi xaXdcrdcoaav. 

8'. Kai xo [Lev duo xrjc; Ttpoxe'fkiaric; EU'ddat; xexpdyo- 
vov prjxov, xal xd xouxw au^erxpa p/]xd, xd 5e xouxw 
dau[i^i£xpa aXoya xaXeio'dw, xal ai 5uvd^evai auxd dXoyoi, 
el [lev xexpdywva d'r), auxal ai TtXerupai, el 8e exepd xiva 
eu-duypa^ia, ai laa auxoTc xexpdywva dvaypdcpouaai. 



Definitions 

1. Those magnitudes measured by the same measure 
are said (to be) commensurable, but (those) of which no 
(magnitude) admits to be a common measure (are said 
to be) incommensurable. t 

2. (Two) straight-lines are commensurable in square* 
when the squares on them are measured by the same 
area, but (are) incommensurable (in square) when no 
area admits to be a common measure of the squares on 
them. § 

3. These things being assumed, it is proved that there 
exist an infinite multitude of straight-lines commensu- 
rable and incommensurable with an assigned straight- 
line — those (incommensurable) in length only, and those 
also (commensurable or incommensurable) in squared 
Therefore, let the assigned straight-line be called ratio- 
nal. And (let) the (straight-lines) commensurable with it, 
either in length and square, or in square only, (also be 
called) rational. But let the (straight-lines) incommensu- 
rable with it be called irrational.* 

4. And let the square on the assigned straight-line be 
called rational. And (let areas) commensurable with it 
(also be called) rational. But (let areas) incommensu- 
rable with it (be called) irrational, and (let) their square- 
roots* (also be called) irrational — the sides themselves, if 
the (areas) are squares, and the (straight-lines) describ- 
ing squares equal to them, if the (areas) are some other 
rectilinear (figure) J 



t In other words, two magnitudes a and l3 are commensurable if a : (3 :: 1 : k, and incommensurable otherwise, 
t Literally, "in power". 

§ In other words, two straight-lines of length a and are commensurable in square if a : (3 :: 1 : A; 1 / 2 , and incommensurable in square otherwise. 
Likewise, the straight-lines are commensurable in length if a : /3 :: 1 : k, and incommensurable in length otherwise. 

11 To be more exact, straight-lines can either be commensurable in square only, incommensurable in length only, or commenusrable/incommensurable 
in both length and square, with an assigned straight-line. 

* Let the length of the assigned straight-line be unity. Then rational straight-lines have lengths expressible as k or A; 1 / 2 , depending on whether 
the lengths are commensurable in length, or in square only, respectively, with unity. All other straight-lines are irrational. 
$ The square-root of an area is the length of the side of an equal area square. 

II The area of the square on the assigned straight-line is unity. Rational areas are expressible as A. All other areas are irrational. Thus, squares 
whose sides are of rational length have rational areas, and vice versa. 



a. . 

Auo [iEyE'dcSv dviawv exxei|jiev«v, edv duo xou y.eiZ,ovoz 
dcpaipsiS/j \±eXt,o\ f\ xo rj^iau xal xou xaxaXsurojievou \±elZ,o\i 
fj xo fjjjiiau, xal xouxo del yiyvrjxai, XeicpiD^asxal xi \ieye , doc;, 
o eaxai eXaoaov xou exxa^ievou eXdaaovo<; \ieye-Qouc,. 

"Eax« 8uo [leye'dr] aviaa xd AB, T, Sv [Le%ov xo AB- 



Proposition V 

If, from the greater of two unequal magnitudes 
(which are) laid out, (a part) greater than half is sub- 
tracted, and (if from) the remainder (a part) greater than 
half (is subtracted), and (if) this happens continually, 
then some magnitude will (eventually) be left which will 



282 



ETOIXEIftN i'. 



ELEMENTS BOOK 10 



Xeyw, on, eav duo xou AB dcpaipcdfj [icTi^ov fj to fjjiiau 
xdi xoO xaxaXemo^tevou ueTCov fj to fj^iiau, xdi xouxo del 
y LYVY]Tai, XeicpiJfiaexai tl ^eyei!}oc;, o eaxai eXaaaov xou T 
^eyei&ouc;. 



AK B 



be less than the lesser laid out magnitude. 

Let AB and C be two unequal magnitudes, of which 
(let) AB (be) the greater. I say that if (a part) greater 
than half is subtracted from AB, and (if a part) greater 
than half (is subtracted) from the remainder, and (if) this 
happens continually, then some magnitude will (eventu- 
ally) be left which will be less than the magnitude C. 

A K H B 



r' ' 

i 1 1 1 

A Z H E 

To T ydp TioXXaTtXaaiaCo^evov eaxai Ttoxe xou AB 
(jiEiCov. TieTioXXomXaaidcTdw, xdi eaxw xo AE xou usv T 
TtoXXarcXaaiov, xou Be AB jieli^ov, xdi Sirjpfja'dco xo AE eic, 
xd ifi T iaa xd AZ, ZH, HE, xdi dcprjprjadco drco \xe\ xou 
AB [isi^ov fj xo f)(iiou xo BO, duo Be xou AO jjieT^ov fj xo 
fj[iiau xo OK, xdi xouxo del yiyveadio, ewe; dv di ev xai AB 
Biaipeaeu; loo7xX/)i9eTc yevcovxai xau; ev xcp AE Biaipeaeaiv. 

"Eaxwaav ouv di AK, KO, OB Biaipeaeu; laonXiQ'&eTe; 
ouaai xau; AZ, ZH, HE - xdi cuel (lel^ov eaxi xo AE xou 
AB, xdi dcpf]pr]xai dno jiev xou AE eXaaaov xou f](iiaeco<; xo 
EH, aTio Be xou AB [lei^ov fj xo fj^iiau xo BO, Xoitiov dpa 
xo HA Xoitcou xou OA jieT^ov eaxiv. xdi enel jiel^ov eaxi xo 
HA xou OA, xdi dcpf)pr]xai xou [iev HA fj|jiiau xo HZ, xou 
Be OA (icT^ov fj xo fjjiiau xo OK, Xomov dpa xo AZ Xoittou 
xou AK jiel^ov eaxiv. Taov Be xo AZ x£i r - xdi xo T dpa 
xou AK [icT^ov eoxiv. eXaaaov dpa xo AK xou T. 

KaxaXeinexai dpa duo xou AB jieye-douc; xo AK (leyeOoc; 
eXaaaov dv xou exxeijjievou eXdaaovoc (icye'dou^ xou E 
onep eBei BeT^ai. — o^ioimc; Be Beix^fpexai, xav f][iiar] fj xd 
dcpaipou^ieva. 



D F G E 

For C, when multiplied (by some number), will some- 
times be greater than AB [Def. 5.4]. Let it have been 
(so) multiplied. And let DE be (both) a multiple of C, 
and greater than AB. And let DE have been divided into 
the (divisions) DF, FG, GE, equal to C. And let BH, 
(which is) greater than half, have been subtracted from 
AB. And (let) HK, (which is) greater than half, (have 
been subtracted) from AH. And let this happen continu- 
ally, until the divisions in AB become equal in number to 
the divisions in DE. 

Therefore, let the divisions (in AB) be AK, KH, HB, 
being equal in number to DF, FG, GE. And since DE is 
greater than AB, and EG, (which is) less than half, has 
been subtracted from DE, and BH, (which is) greater 
than half, from AB, the remainder GD is thus greater 
than the remainder HA. And since GD is greater than 
HA, and the half GF has been subtracted from GD, and 
HK, (which is) greater than half, from HA, the remain- 
der DF is thus greater than the remainder AK. And DF 
(is) equal to C. C is thus also greater than AK. Thus, 
AK (is) less than C. 

Thus, the magnitude AK, which is less than the lesser 
laid out magnitude C, is left over from the magnitude 
AB. (Which is) the very thing it was required to show. — 
(The theorem) can similarly be proved even if the (parts) 
subtracted are halves. 



t This theorem is the basis of the so-called method of exhaustion, and is generally attributed to Eudoxus of Cnidus. 



P'- 

'Edv Buo ^eyei9Gv [exxeijieviov] dviawv dvducpaipou^ievou 
del xou eXdaaovot; duo xou ^.ei^ovoc; xo xaxaXeiTio^ievov 
^iT)8e7ioxe xaxa^texpfj xo npo eauxou, dau^expa eaxai xd 
^eyc'dr). 

Auo yap jieye-daiv ovxtov dviawv xwv AB, TA xdi 
eXdaaovoc; xou AB dvducpaipou^ievou del xou eXdaaovoc; 
duo xou [iei^ovoc; xo nepiXemojievov [irjBejraxe xaxa^e- 



Proposition 2 

If the remainder of two unequal magnitudes (which 
are) [laid out] never measures the (magnitude) before it, 
(when) the lesser (magnitude is) continually subtracted 
in turn from the greater, then the (original) magnitudes 
will be incommensurable. 

For, AB and CD being two unequal magnitudes, and 
AB (being) the lesser, let the remainder never measure 



283 



ETOIXEIftN i'. 



ELEMENTS BOOK 10 



xpeixo to Tipo eauxou- Xeyw, oxi dauuuexpd eaxi xd AB, 
TA ueye'dr). 

AH B 

i — i 1 1 

E' ' 

i 1 1 1 

r z a 

El ydp laxi auuuexpa, uexprpei xi auxa ueycdo<;. ue- 
xpeixw, ei 8uvax6v, xod eaxw xo E- xal xo uev AB xo ZA 
xaxauexpouv XeiTiexo eauxou eXaaaov xo TZ, xo 8e TZ xo 
BH xaxauexpouv Xeircexo eauxou eXaaaov xo AH, xal xouxo 
del yivecrdco, ecog ou Xeicp^fj xi ueycdoc;, 6 eaxiv eXaaaov xou 
E. yeyovexco, xal XsXsicp'Ow xo AH eXaaaov xou E. insi ouv 
xo E xo AB uexpeT, dXXd xo AB xo AZ uexpel, xai xo E apa 
xo ZA uexpr|aei. uexpeT 8e xal 6Xov xo TA- xal Xomov apa 
xo rZ uexpiqaei. dXXd xo TZ xo BH uexpeT- xal xo E apa 
xo BH uexpeT. uexpeT 8e xal oXov xo AB- xal Xomov apa xo 
AH uexprpei, xo ueT^ov xo eXaaaov onep eaxlv dSuvaxov. 
oux apa xd AB, TA ueye'dr] uexprpei xi ueyeiSoc;- dauuuexpa 
apa eaxl xd AB, FA ueyeiSr). 

'Edv apa 8uo ueyei!)Gv dviaiov, xal xd e^fjg. 

t The fact that this will eventually occur is guaranteed by Prop. 10.1. 



the (magnitude) before it, (when) the lesser (magnitude 
is) continually subtracted in turn from the greater. I say 
that the magnitudes AB and CD are incommensurable. 

A G B 

i 1 1 1 

Ei 1 



C F D 

For if they are commensurable then some magnitude 
will measure them (both). If possible, let it (so) measure 
(them), and let it be E. And let AB leave CF less than 
itself (in) measuring FD, and let CF leave AG less than 
itself (in) measuring BG, and let this happen continually, 
until some magnitude which is less than E is left. Let 
(this) have occurred, ^ and let AG, (which is) less than 
E, have been left. Therefore, since E measures AB, but 
AB measures DF, E will thus also measure FD. And it 
also measures the whole (of) CD. Thus, it will also mea- 
sure the remainder CF. But, CF measures BG. Thus, E 
also measures BG. And it also measures the whole (of) 
AB. Thus, it will also measure the remainder AG, the 
greater (measuring) the lesser. The very thing is impos- 
sible. Thus, some magnitude cannot measure (both) the 
magnitudes AB and CD. Thus, the magnitudes AB and 
CD are incommensurable [Def. 10.1]. 

Thus, if ... of two unequal magnitudes, and so on 



y'. Proposition 3 

Auo ueycdcov auuuexpcov Boftevxcov xo ueyiaxov auxfiv To find the greatest common measure of two given 

xoivov uexpov eupeTv. commensurable magnitudes. 

AZ B A F B 

i — i — i — i — i i— i — i — i — i 



r E ACE D 



Hi ' 

'Eaxw xd 5oi9evxa 8uo ueye-dr) auuuexpa xd AB, TA, 
Sv eXaaaov xo AB- Bel 8f) xov AB, TA xo ueyiaxov xoivov 
uexpov eupeTv. 

To AB ydp uiycdoc; rjxoi uexpeT xo TA fj ou. et uev 
ouv uexpel, uexpel 8e xal eauxo, xo AB apa xfiv AB, TA 
xoivov uexpov eaxiv xal cpavepov, oxi xal ueyiaxov. ueTCov 
ydp xou AB ueyei!)ou<; xo AB ou uexprpei. 

Mr) uexpeixw 5r] xo AB xo FA. xal dvducpaipouuevou 
del xou eXdaaovoc; duo xou ueiCovoc;, xo TtepiXeiitouevov 
uexprpei Ttoxe xo Tipo eauxou 8ia xo \ir) elvai dauuuexpa xd 
AB, TA- xal xo uev AB xo EA xaxauexpouv Xcikcxw eauxou 



Gi ' 

Let AB and CD be the two given magnitudes, of 
which (let) AB (be) the lesser. So, it is required to find 
the greatest common measure of AB and CD. 

For the magnitude AB either measures, or (does) not 
(measure), CD. Therefore, if it measures (CD), and 
(since) it also measures itself, AB is thus a common mea- 
sure of AB and CD. And (it is) clear that (it is) also (the) 
greatest. For a (magnitude) greater than magnitude AB 
cannot measure AB. 

So let AB not measure CD. And continually subtract- 
ing in turn the lesser (magnitude) from the greater, the 



284 



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sXaaaov to Er, to 8e Er to ZB xaxa^iexpouv Xemexo 
EauxoD eXaaaov xo AZ, xo 8e AZ xo EE ^texpeixw. 

Tkei ouv xo AZ xo EE [iexpei, dXXa xo EE xo ZB jxexpel, 
xal xo AZ apa xo ZB ^exprjaet. jiexpei 8e xal eauxo' xal 
oXov apa xo AB ^exprpei xo AZ. dXXa xo AB xo AE [lexpsi- 
xal xo AZ apa xo EA ^expiqoei. piexpei 8e xal xo EE- xai 
oXov apa xo TA [isxpel' xo AZ apa xaiv AB, EA xoivov 
^.fxpov eaxiv. Xeyw 8Vj, oxi xal jieyiaxov. £ i Y<*P t 1 ^ eaxai 
xi [isys'doc jiei£ov xou AZ, o ^sxpfjaei xd AB, EA. eax« xo 
H. excel ouv xo H xo AB ^.sxpei, dXXa xo AB xo EA (lexpsi, 
xal xo H apa xo EA ^exprjaei. ^lExpeT Se xal oXov xo EA- 
xal Xomov apa xo TE [lexprpei xo H. dXXa xo TE xo ZB 
^.expsT' xal xo H apa xo ZB jiexprjaei. ^.expel 8e xal oXov 
xo AB, xal Xomov xo AZ (jiexprpei, xo ^.el^ov xo IXaaaov 
oTiep eaxlv dSuvaxov. oux apa ^leT^ov xi [leye-Qoc, xou AZ 
xd AB, EA (jiexpr]0£i- xo AZ apa xo5v AB, EA xo jieyiaxov 
xoivov ^.expov scrav. 

Auo apa [isyeiSGv au^iexptov Bo-dcvxcov x£Sv AB, EA 
xo ^icyiaxov xoivov jiexpov r]upr)xai- onsp cBei Ba^ai. 



remaining (magnitude) will (at) some time measure the 
(magnitude) before it, on account of AB and CD not be- 
ing incommensurable [Prop. 10.2]. And let AB leave EC 
less than itself (in) measuring ED, and let EC leave AF 
less than itself (in) measuring FB, and let AF measure 
CE. 

Therefore, since AF measures CE, but CE measures 
FB, AF will thus also measure FB. And it also mea- 
sures itself. Thus, AF will also measure the whole (of) 
AB. But, AB measures DE. Thus, AF will also mea- 
sure ED. And it also measures CE. Thus, it also mea- 
sures the whole of CD. Thus, AF is a common measure 
of AB and CD. So I say that (it is) also (the) greatest 
(common measure) . For, if not, there will be some mag- 
nitude, greater than AF, which will measure (both) AB 
and CD. Let it be G. Therefore, since G measures AB, 
but AB measures ED, G will thus also measure ED. And 
it also measures the whole of CD. Thus, G will also mea- 
sure the remainder CE. But CE measures FB. Thus, G 
will also measure FB. And it also measures the whole 
(of) AB. And (so) it will measure the remainder AF, 
the greater (measuring) the lesser. The very thing is im- 
possible. Thus, some magnitude greater than AF cannot 
measure (both) AB and CD. Thus, AF is the greatest 
common measure of AB and CD. 

Thus, the greatest common measure of two given 
commensurable magnitudes, AB and CD, has been 
found. (Which is) the very thing it was required to show. 



H6pio\J.a. 

'Ex 6r| xouxou cpavepov, oxi, edv [leye-Qoc, Suo jisys-dr] 
tiexprj, xal xo pieyiaxov auxwv xoivov ^expov [isxprpsi. 



Corollary 

So (it is) clear, from this, that if a magnitude measures 
two magnitudes then it will also measure their greatest 
common measure. 



5'. 

Tpiwv ^leye'dcov au^iexpcov SoiSsvxwv xo [aeyioTov 
auxfiv xoivov jiexpov eupsTv. 

A' 1 

B 1 

r> 1 

i — i i — i i — i 

A E Z 

TSaxw xd SoiJevxa xpia ^leye'dr] au^erxpa xd A, B, E- 
8eT 8/) xfiv A, B, T xo [isyiaxov xoivov jisxpov supeiv. 

EiXrjcp'dto yap 8uo x«v A, B xo [isyiaxov xoivov ^xsxpov, 
xal screw xo A- xo 8/) A xo T fjxoi jiexpsl fj ou [jiexpsT]. 
^expeixw itpoxepov. enel ouv xo A xo E [isxpei, [isxpeT Se 



Proposition 4 

To find the greatest common measure of three given 
commensurable magnitudes. 

A i 1 

B' ' 

C' 1 

i 1 i — i i 1 

D E F 

Let A, B, C be the three given commensurable mag- 
nitudes. So it is required to find the greatest common 
measure of A, B, C. 

For let the greatest common measure of the two (mag- 
nitudes) A and B have been taken [Prop. 10.3], and let it 



285 



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ELEMENTS BOOK 10 



xal xd A, B, to A dpa xd A, B, T [lexpel/ xo A apa xwv A, 
B, r xoivov [iexpov eaxiv. xal cpavepov, oxi xal [icyiaxov 
[iclCov yap xou A [ieye , dou<; xa A, B ou [lexpel. 

Mr) [icxpeixo Sr] xo A xo r. Xey« xpGxov, oxi au[i[iexpd 
eaxi xa r, A. exel yap au[i[iexpd eaxi xa A, B, T, [iexpr]aei 
xi auxd [ieycdo<;, o Sr]XaSr) xal xa A, B [iexprjaei- &axe 
xal xo iwv A, B [icyiaxov xoivov [iexpov xo A [iexprjaei. 
[icxpeT Be xal xo E &axe xo eipr][ievov [icycdoc; [iexprjaei xa 
r, A- au[i[iexpa apa eaxl xa T, A. elXrjcp'dco ouv auxGv xo 
[icyiaxov xoivov [iexpov, xal eaxw xo E. exel ouv xo E xo 
A [iexpeT, dXXd xo A xd A, B [iexpeT, xal xo E dpa xa A, B 
[lexprpei. [iexpeT Se xal xo T. xo E dpa xd A, B, T [iexpeT/ 
xo E apa xwv A, B, T xoivov eaxi (iexpov. Xeyw oxi xal 
[icyiaxov. ei yap Buvaxov, eaxw xi xou E [iel£ov [leycdot; 
xo Z, xal [icxpeixw xd A, B, T. xal exel xo Z xd A, B, T 
[iexpeT, xal xd A, B apa (iexprjaei xal xo xov A, B [icyiaxov 
xoivov [iexpov [lexprpei. xo 8e xwv A, B (icyiaxov xoivov 
[iexpov eaxl xo A- xo Z apa xo A [iexpeT. [iexpeT Se xal xo 
r- xo Z apa xd T, A [iexpeT' xal xo xQv T, A apa [icyiaxov 
xoivov (iexpov [iexprjaei xo Z. eaxi 8e xo E- xo Z apa xo E 
[iexprjaei, xo [leT^ov xo eXaaaov oxep eaxlv dSuvaxov. oux 
apa [ieT£6v xi xou E [icye-doug [[leycdoc] xd A, B, T [iexpeT- 
xo E apa xwv A, B, T xo [icyiaxov xoivov [iexpov eaxlv, edv 
[ir] [icxpfj xo A xo T, edv 8e [icxpfj, auxo xo A. 

Tpiwv dpa [leyeiDwv au[i[iexp«v So'devxwv xo [icyiaxov 
xoivov [iexpov rjuprjxai [oxep e8ei Bel^ai]. 



be D. So D either measures, or [does] not [measure], C. 
Let it, first of all, measure (C). Therefore, since D mea- 
sures C, and it also measures A and B, D thus measures 

A, B, C. Thus, D is a common measure of A, B, C. And 
(it is) clear that (it is) also (the) greatest (common mea- 
sure). For no magnitude larger than D measures (both) 
A and B. 

So let D not measure C. I say, first, that C and D are 
commensurable. For if A, B, C are commensurable then 
some magnitude will measure them which will clearly 
also measure A and B. Hence, it will also measure D, the 
greatest common measure of A and B [Prop. 10.3 corr.]. 
And it also measures C. Hence, the aforementioned mag- 
nitude will measure (both) C and D. Thus, C and D are 
commensurable [Def. 10.1]. Therefore, let their greatest 
common measure have been taken [Prop. 10.3], and let 
it be E. Therefore, since E measures D, but D measures 
(both) A and B, E will thus also measure A and B. And 
it also measures C. Thus, E measures A, B, C. Thus, E 
is a common measure of A, B, C. So I say that (it is) also 
(the) greatest (common measure) . For, if possible, let F 
be some magnitude greater than E, and let it measure A, 

B, C. And since F measures A, B, C, it will thus also 
measure A and B, and will (thus) measure the greatest 
common measure of A and B [Prop. 10.3 corr.]. And D 
is the greatest common measure of A and B. Thus, F 
measures D. And it also measures C. Thus, F measures 
(both) C and D. Thus, F will also measure the greatest 
common measure of C and D [Prop. 10.3 corr.]. And it is 
E. Thus, F will measure E, the greater (measuring) the 
lesser. The very thing is impossible. Thus, some [magni- 
tude] greater than the magnitude E cannot measure A, 
B, C. Thus, if D does not measure C then E is the great- 
est common measure of A, B, C. And if it does measure 
(C) then D itself (is the greatest common measure). 

Thus, the greatest common measure of three given 
commensurable magnitudes has been found. [(Which is) 
the very thing it was required to show] 



H6pio\J.a. 

'Ex 8r| xouxou (pavepov, oxi, edv [icye'doc; xpia (leye'dr) 
[lexpfj, xal xo [icyiaxov auxaiv xoivov [iexpov [iexprjaei. 

! 0\±oiu>q Srj xal exl xXeiovwv xo [icyiaxov xoivov [iexpov 
Xrjcp-drjaexai, xal xo xopia[ia xpo)(iopr]aei. onep e8ei Sel^ai. 



Corollary 

So (it is) clear, from this, that if a magnitude measures 
three magnitudes then it will also measure their greatest 
common measure. 

So, similarly, the greatest common measure of more 
(magnitudes) can also be taken, and the (above) corol- 
lary will go forward. (Which is) the very thing it was 
required to show. 



286 



ETOIXEIftN i'. 



ELEMENTS BOOK 10 



z. Proposition 5 

Td atiuuexpa jieye'v}/] npbc, dXX/jXa Xoyov e/ei, ov Commensurable magnitudes have to one another the 
apiduoc; npbc, dpidfjiov. ratio which (some) number (has) to (some) number. 

A BT A B C 

i — i — i — i i — i — i i — i i — i — i — i i — i — i i — i 



"Eaxw auu^iexpa ^leye'dr] xa A, B- Xeyto, oxi to A Ttpoc 
to B Xoyov £X £l > ° v otpi'd^ioc; npbq dpid^iov. 

'Enel yap ouujiSTpd eaxi Ta A, B, jiexprpei ti auxa 
[isys-Qoz. jiexpeixw, xal eaxio to T. xal oadxu; to T to 
A [lexpei, xoaauxai uovdBec eaxoaav ev xG A, oadxu; Be: 
to r to B [lexpel, xoaauxai uovdSec; eaxcoaav ev xa> E. 

Tkel ouv to T to A jiexpeT xaxd xac sv tw A uovd8a<;, 
^texpeT Be xal f] uovac; xov A xaxd xa<; ev auxw ^ovdSac, 
ladxu; dpa f] ^tovac; tov A jiexpeT api'duov xal to T uey^ ? 
to A- eaxiv dpa w<; to T Ttpoc; to A, outck f] [Lovaz npbc, 
tov A- dvduaXiv dpa, cb? to A Tip6<; to T, oux«<; 6 A icpoc; 
t/]v ^.ovdBa. udXiv eirel to T to B ^expel xaxd xac; ev iu 
E ^lovdSac;, ^texpeT 8e xal f] uovdc; tov E xaTa xac; ev auxcp 
uovd8a<;, ladxu; dpa f] jxovac; tov E ^texpeT xal to T to B' 
eaTiv dpa w<; to T npbz to B, ouxmc; f) jiovdc; icpoc; tov E. 
eBelx'dr) 8e xal cbc; to A Tcpoc; to T, 6 A Tcpoc; t/]v jiovdBa- 
8i° laou dpa eaTiv cbc; to A jcp6<; to B, ouxcoc; 6 A apiduoc; 
Ttp6<; tov E. 

Ta dpa au[i^expa ueye'dr) Ta A, B Tcpoc; aXXrjXa Xoyov 
exei, Sv dpnf)[i6<; 6 A icpoc dpi%6v tov E- oicep eBei SeT^ai. 



Let A and B be commensurable magnitudes. I say 
that A has to B the ratio which (some) number (has) to 
(some) number. 

For if A and B are commensurable (magnitudes) then 
some magnitude will measure them. Let it (so) measure 
(them), and let it be C. And as many times as C measures 

A, so many units let there be in D. And as many times as 
C measures B, so many units let there be in E. 

Therefore, since C measures A according to the units 
in D, and a unit also measures D according to the units 
in it, a unit thus measures the number D as many times 
as the magnitude C (measures) A. Thus, as C is to A, 
so a unit (is) to D [Dei. 7. 20] J Thus, inversely, as A (is) 
to C, so D (is) to a unit [Prop. 5.7 corr.]. Again, since 
C measures B according to the units in E, and a unit 
also measures E according to the units in it, a unit thus 
measures E the same number of times that C (measures) 

B. Thus, as C is to B, so a unit (is) to E [Def. 7.20]. And 
it was also shown that as A (is) to C, so D (is) to a unit. 
Thus, via equality, as A is to B, so the number D (is) to 
the (number) E [Prop. 5.22]. 

Thus, the commensurable magnitudes A and B have 
to one another the ratio which the number D (has) to the 
number E. (Which is) the very thing it was required to 
show. 



t There is a slight logical gap here, since Def. 7.20 applies to four numbers, rather than two number and two magnitudes. 



Edv 860 [leye'f)/) Ttpoc; dXXrjXa Xoyov eyjr), ov dpidjjidc; 
Ttpoc; dpi%6v, au^expa eaxai Ta [leye'dr]. 

A 1 — 1 — 1 — 1 — 1 B 1 1 

A' ' E' 1 

pi — 1 Z' — 1 — 1 — 1 

Auo yap fieye'dr] xd A, B Ttpoc; dXXr)Xa Xoyov e^exco, ov 
dpiOjjiog 6 A Ttpoc; dprd^iov tov E- Xeyw, 6x1 atiunexpa eaxi 
Ta A, B ^eyei}/]. 

"Oaai yap eicfiv ev tu A ^tovdSec;, ek ToaaOTa laa 



Proposition 6 

If two magnitudes have to one another the ratio which 
(some) number (has) to (some) number then the magni- 
tudes will be commensurable. 

A I — I — I — I 1 B 1 1 

D' 1 E i 1 

C 1 — 1 F 1 — i — i — 1 

For let the two magnitudes A and B have to one an- 
other the ratio which the number D (has) to the number 
E. I say that the magnitudes A and B are commensu- 
rable. 



287 



ETOIXEIfiN i'. 



ELEMENTS BOOK 10 



SirpiqcrdM to A, xal evi auTGv Taov sgtw to r - oaai 6e 
rimv sv to E [iovd8ec, ex toooutmv [icycdGv lacov to T 
auyxeiai&M to Z. 

'EtieI ouv, oaai eioiv ev tG A (iovd8ec, ToaauTa rim xai 
ev tG A ^eyedr) ^ aa T Q r, o dpa jiepoc taxiv f] jiovac tou 
A, to auTo [izpoc, ecru xal to T tou A- eaTiv apa Gc to T 
Tipoc to A, outgjc f) jiovac Tipoc tov A. 8s f) (iovac 

tov A dpiOjiov (iETpsI dpa xal to V to A. xal ene'i eaTiv 
Gc to T Tipoc to A, outioc f) ^ovdc Ttpoc tov A [dpidjiov], 
dvaTiaXiv apa Gc to A Ttpoc to T, outwc 6 A dpidjioc Ttpoc 
tt)v [iovd8a. TtdXiv insi, oaai eialv ev ifi E [iovdSec, ToaauTa 
eloi xal ev tG Z I'oa tG T, eaTiv apa Gc to T Tipoc to Z, 
outcoc f) ^xovac Ttpoc tov E [dpi%6v]. eSei/i}/) 8e xal Gc 
to A Ttpoc to r, outcoc 6 A Ttpoc tt)v [iovd5a' 5i' laou apa 
cotIv Gc to A Ttpoc to Z, outoc 6 A Ttpoc tov E. dXX'' Gc 6 
A Ttpoc tov E, outw<; cotI to A Ttpoc to B - xal Gc dpa to A 
Ttpoc to B, outwc xal Ttpoc to Z. to A apa Ttpoc exaTcpov 
tGv B, Z tov ai)Tov §)(ei Xoyov I'aov apa eaTi to B tG Z. 
[iCTpel 8e to T to Z - ^.eTpeT dpa xal to B. dXXd \ir\M xal to 
A' to r dpa Ta A, B [iCTpeT. aujijiCTpov dpa sotI to A tG 
B. 

'Eav dpa 8uo (jieyeiSr) Tipoc dXXrjXa, xal Ta ec"rjc- 



LTopiajJia. 

'Ex 8r) toutou cpavepov, oti, eav Gai 8uo dpi , d|ioi, Gc 
oi A, E, xal suiDeTa, Gc f) A, SuvaTov eaTi Tioifjaai Gc 6 
A dpid^ioc Ttpoc tov E dpi/d^iov, outoc ttjv eu-delav Tipoc 
euif)eTav. eav Se xal tGv A, Z [Leaf] dvdXoyov Xrjcp'dfj, Gc rj 
B, soTai Gc f) A Ttpoc; t?]v Z, outoc to dito xrjc; A Ttpoc; to 
aTco Tfjc B, touteotiv Gc f) TcpGTT) Ttpoc t/]v Tp[T/]v, outoc 
to aTco Tfjc TtpGTTjc Ttpoc; to aTto Tfjc BeuTcpac to ojioiov xal 
o^toiwc dvaypacpo^tevov. dXX' Gc; f\ A Ttpoc t/]v Z, outok 
eaTiv 6 A dpid^oc; upoc; tov E dpn!)^6v ysyovev dpa xal 
Gc; 6 A dpiiD^toc; Tipoc; tov E dpnD^tov, outwc; to dito Tfjc A 
eu-delac Ttpoc to duo Tfjc B eu^eiac oTcep e8ei 8ric;ai. 



Ta douji|jieTpa ^ieyei9r) Tipoc dXXr)Xa Xoyov oux ex ei J ° v 
dpn!)^6c Tipoc dpidjiov. 

'EaTW dau^i^ieTpa ^.eye'dr] Ta A, B- Xeyco, oti to A Tipoc 
to B Xoyov oux exei, 8v dpi-djioc Tipoc dpid^tov. 



For, as many units as there are in D, let A have been 
divided into so many equal (divisions). And let C be 
equal to one of them. And as many units as there are 
in E, let F be the sum of so many magnitudes equal to 
C. 

Therefore, since as many units as there are in D, so 
many magnitudes equal to C are also in A, therefore 
whichever part a unit is of D, C is also the same part of 
A. Thus, as C is to A, so a unit (is) to D [Def. 7.20]. And 
a unit measures the number D. Thus, C also measures 
A. And since as C is to A, so a unit (is) to the [number] 
D, thus, inversely, as A (is) to C, so the number D (is) 
to a unit [Prop. 5.7 corr.]. Again, since as many units as 
there are in E, so many (magnitudes) equal to C are also 
in F, thus as C is to F, so a unit (is) to the [number] E 
[Def. 7.20]. And it was also shown that as A (is) to C, 
so D (is) to a unit. Thus, via equality, as A is to F, so D 
(is) to E [Prop. 5.22]. But, as D (is) to E, so A is to B. 
And thus as A (is) to B, so (it) also is to F [Prop. 5.11]. 
Thus, A has the same ratio to each of B and F. Thus, i? is 
equal to F [Prop. 5.9]. And C measures F. Thus, it also 
measures B. But, in fact, (it) also (measures) A. Thus, 
C measures (both) A and B. Thus, ^4 is commensurable 
withB [Def. 10.1]. 

Thus, if two magnitudes ... to one another, and so on 



Corollary 

So it is clear, from this, that if there are two numbers, 
like D and E, and a straight-line, like A, then it is possible 
to contrive that as the number D (is) to the number E, 
so the straight-line (is) to (another) straight-line (i.e., F). 
And if the mean proportion, (say) B, is taken of A and 
F, then as A is to F, so the (square) on A (will be) to the 
(square) on B. That is to say, as the first (is) to the third, 
so the (figure) on the first (is) to the similar, and similarly 
described, (figure) on the second [Prop. 6.19 corr.]. But, 
as A (is) to F, so the number D is to the number E. Thus, 
it has also been contrived that as the number D (is) to 
the number E, so the (figure) on the straight-line A (is) 
to the (similar figure) on the straight-line B. (Which is) 
the very thing it was required to show. 

Proposition 7 

Incommensurable magnitudes do not have to one an- 
other the ratio which (some) number (has) to (some) 
number. 

Let A and B be incommensurable magnitudes. I say 
that A does not have to B the ratio which (some) number 
(has) to (some) number. 



288 



ETOIXEIftN i'. 



ELEMENTS BOOK 10 



A' ' 

B 1 

El yap zyzi xo A npo<; to B Xoyov, ov dtpi , d(Ji6c; upoc; 
dpi%6v, auji^texpov eaxai xo A iw B. oux eoxi Be- oux apa 
xo A 7ip6<; xo B Xoyov zyzi, ov apiduoc; npoz dpid^ov. 

Td apa dau^[iexpa ^teys'dr] 7ipo<; aXXrjXa Xoyov oux e/ei, 
xal xd eZ>y]z- 

*)'• 

'Edv 8uo (ieyeiSr) npbc, dXX/]Xa Xoyov [ir\ z^f], 8v dpi'd^oc; 
npbc, apiduov, dau^iexpa eaxai xa ^eyei&r]. 

A' ' 

B 1 

Auo yap ^xsyei}/) xa A, B Tipoc; aXXrjXa Xoyov [ir] E)(£x(o, 
ov dpi'djj.ot; upog apidjjiov Xeyw, oxi dau^expd eaxi xa A, 
B ^sysdr). 

Ei yap Eaxai aujijiexpa, xo A upog xo B Xoyov zZ